MATHS M.C.Q (1 Marks)

3. $\frac{\pi}{6}$

Solution:

Given:

$\sqrt{3}(\cot\text{x}+\tan\text{x})=4$

$\Rightarrow\sqrt{3}\Big(\frac{\cos\text{x}}{\sin\text{x}}+\frac{\sin\text{x}}{\cos\text{x}}\Big)=4$

$\Rightarrow\sqrt{3}(\cos^2\text{x}+\sin^2\text{x})=4\sin\text{x}\cos\text{x}$

$\Rightarrow\sqrt{3}=2\sin2\text{x}$ $\big[\sin2\text{x}=2\sin\text{x}\cos\text{x}\big]$

$\Rightarrow\sin2\text{x}=\frac{\sqrt{3}}{2}$

$\Rightarrow\sin2\text{x}=\sin\frac{\pi}{3}$

$\Rightarrow2\text{x}=\text{n}\pi+(-1)^{\text{n}}\frac{\pi}{3},\text{n}\in\text{Z}$

$\Rightarrow\text{x}=\frac{\text{n}\pi}{2}+(-1)^{\text{n}}\frac{\pi}{6},\text{n}\in\text{Z}$

To obtain the smallest value of x, we will put n = 0n = 0 in the above equation.

Thus, we have:

$\text{x}=\frac{\pi}{6}$

Hence, the smallest value of x is $\frac{\pi}{6}.$

3.  $\text{x}=\frac{\text{n}\pi}{7}+\frac{\pi}{14},\ \text{n}\in\text{Z}$

Solution:

Given:

$\tan5\text{x}=\cot2\text{x}$

$\Rightarrow\tan5​\text{x}=\tan\Big(\frac{\pi}{2}-2\text{x}\Big)$

$\Rightarrow5\text{x}=\text{n}\pi+\frac{\pi}{2}-2\text{x}$

$\Rightarrow7\text{x}=\text{n}\pi+\frac{\pi}{2}$

$\Rightarrow\text{x}=\frac{\text{n}\pi}{7}+\frac{\pi}{14},\ \text{n}\in\text{Z}$ 

2. $\text{x}=\text{n}\pi\pm\frac{\pi}{3},\ \text{n}\in\text{Z}$

Solution:

Given:

$\sqrt{3}\sin​\text{x}+\cos\text{x}=\sqrt{3}\ .....(1)$

This equation is of the form $\text{a}\sin\theta+\text{b}\cos\theta=\text{c},$ where $\text{a}=\sqrt{3},\ \text{b}=1$ and $\text{c}=\sqrt{3}$

Let:

$\text{a}=\text{r}\cos\alpha$ and $\text{b}=\text{r}\sin\alpha$

Now

$\text{r}=\sqrt{\text{a}^2+\text{b}^2}=\sqrt{\Big(\sqrt{3}\Big)^2+1^1}=2$ and $\tan\alpha=\frac{\text{b}}{\text{a}}\Rightarrow\tan\alpha=\frac{1}{\sqrt{3}}$

$\Rightarrow\alpha=\frac{\pi}{6}$

On putting $\text{a}=\sqrt{3}=\text{r}\cos\alpha$ and $\text{b}=1=\text{r}\sin\alpha$ in equation (1),  we get:

$​\text{r}​\cos\alpha\sin​\text{x}​+​\text{r}​\sin\alpha\cos​\text{}x​=\sqrt{3}$

$\Rightarrow\text{r}\sin\big(\text{x}+\alpha\big)=\sqrt{3}$

$\Rightarrow\text{2}\sin\big(\text{x}+\alpha\big)=\sqrt{3}$

$\Rightarrow\text{}\sin\big(\text{x}+\alpha\big)=\frac{\sqrt{3}}{2}$

$\Rightarrow\text{}\sin\big(\text{x}+\alpha\big)=\sin\frac{\pi}{3}$

$\Rightarrow\text{}\sin\big(\text{x}+\frac{\pi}{6}\big)=\sin\frac{\pi}{3}$

$\Rightarrow​​\text{x}=\text{n}\pi+(-1)^\text{n}\frac{\pi}{3}-\frac{\pi}{6},\ \text{n}\in\text{Z}$

1. 1

Solution:

Given:

$\cot\text{x}-\tan\text{x}=\text{a}$

$\Rightarrow\frac{1}{\tan\text{x}}-\tan\text{x}=\text{a}$

$\Rightarrow1-\tan^2\text{x}=\text{a}\tan\text{x}$

$\Rightarrow\tan^2\text{x}+\text{a}\tan\text{x}-1=0$

It is a quadratic equation.

If $\tan\text{x}=\text{z},$ then the equation becomes

$\text{z}^2+\text{az}-1=0$

$\Rightarrow\text{z}=\frac{-\text{a}\pm\sqrt{\text{a}^2+4}}{2}$

$\Rightarrow\tan\text{x}=\frac{-\text{a}\pm\sqrt{\text{a}^2+4}}{2}$

$\Rightarrow\text{x}=\tan^{-1}\Big(\frac{-\text{a}\pm\sqrt{\text{a}^2+4}}{2}\Big)$

There are two roots of the given equation, but we need to find the number of roots in the first quadrant.

There is exactly one root of the equation, that is, $\text{x}=\tan^{-1}\Big(\frac{-\text{a}+\sqrt{\text{a}^2+4}}{2}\Big).$

4. None of these

Solution:

Given equation:

$3\cos\text{x}+4\sin\text{x}=6\ .....(1)$

Thus, the equation is of the form $\text{a}\cos\text{x}+\text{b}\sin\text{x}=\text{c},$ where  and $\text{c}=6.$

Let:

$\text{a}=3=\text{r}\cos\alpha$ and $\text{b}=4=\text{r}\sin\alpha$

Now,

$\tan\alpha=\frac{\text{b}}{\text{a}}=\frac{4}{3}$

$\Rightarrow\alpha=\tan^{-1}\Big(\frac{4}{3}\Big)$

Also,

$\text{r}=\sqrt{\text{a}^2+\text{b}^2}=\sqrt{9+16}=\sqrt{25}=5$

On putting $\text{a}=3=\text{r}\cos\alpha$ and $\text{b}=4=\text{r}\sin\alpha$ in equation (i), we get:

$\Rightarrow\text{r}\cos(\theta-\alpha)=6$

$\Rightarrow\text{5}\cos(\theta-\alpha)=6$

$\Rightarrow\text{}\cos(\theta-\alpha)=\frac{6}{5}$

From here, we cannot find the value of $\theta.$

3. $\text{n}\pi\pm\frac{\pi}{6},\text{n}\in\text{Z}$

Solution:

Given:

$\Rightarrow\sin^2\text{x}=1$

$\Rightarrow\sin^2\text{x}=\frac{1}{4}$

$\Rightarrow\sin\text{x}=\frac{1}{2}$ or $\sin\text{x}=-\frac{1}{2}$

$\Rightarrow\sin\text{x}=\sin\frac{\pi}{6}$ or $\sin\text{x}=\sin\Big(-\frac{\pi}{6}\Big)$

$\Rightarrow\text{x}=\text{n}\pi+(-1)^\text{n}\frac{\pi}{6},\text{n}\in\text{Z}$ or $\text{n}\pi+(-1)^\text{n}\Big(-\frac{\pi}{6}\Big),\text{n}\in\text{Z}$

$\Rightarrow\text{x}=\text{n}\pi\pm\frac{\pi}{6},\text{n}\in\text{Z}$

1. AP

Solution:

Given:

$\tan\text{px}-\tan\text{qx}=0$

$\Rightarrow\tan​​\text{px}=\tan\text{qx}$

$\Rightarrow\frac{\sin\text{px}}{\cos\text{px}}=\frac{\sin\text{qx}}{\cos\text{qx}}$

$\Rightarrow\sin\text{px}\cos\text{qx}=\sin\text{qx}\cos\text{px}$

$\Rightarrow\frac{1}{2}\Big[\sin\Big(\frac{\text{p+q}}{2}\Big)\text{x}+\sin\Big(\frac{\text{p}-\text{q}}{2}\Big)\text{x}\Big]\\=\frac{1}{2}\Big[\sin\Big(\frac{\text{q+p}}{2}\Big)\text{x}+\sin\Big(\frac{\text{q}-\text{p}}{2}\text{x}\Big)\Big]$

Now,

$\Rightarrow\sin\text{A}\cos\text{B}=\frac{1}{2}\Big[\Big(\frac{\text{A+B}}{2}\Big)+\Big(\frac{\text{A}-\text{B}}{2}\Big)\sin\Big]$

$\Rightarrow\sin\Big(\frac{\text{p}-\text{q}}{2}\Big)\text{x}=\sin\Big(\frac{\text{q}-\text{p}}{2}\Big)\text{x}$

$\Rightarrow\sin\Big(\frac{\text{p}-\text{q}}{2}\Big)\text{x}=-\sin\Big(\frac{\text{q}-\text{p}}{2}\Big)\text{x}$

$\Rightarrow2\sin\Big(\frac{​\text{p}-\text{q}​}{2}\Big)​\text{x}=0$

$\Rightarrow\sin\Big(\frac{\text{p}-\text{q}}{2}\Big)\text{x}=0$

$\Rightarrow\Big(\frac{\text{p}-\text{q}}{2}\Big)\text{x}=\text{n}\pi,\text{n}\in\text{Z}$

Now, on putting the value of n, we get:

$\text{n}=1,\text{x}=\frac{2\pi}{(\text{p}-\text{q})}=\text{a}_1$

$\text{n}=2,\text{x}=\frac{4\pi}{(\text{p}-\text{q})}=\text{a}_2$

$\text{n}=3,\text{x}=\frac{6\pi}{(\text{p}-\text{q})}=\text{a}_3$

$\text{n}=4,\text{x}=\frac{4\pi}{(\text{p}-\text{q})}=\text{a}_4$

And so on.

Also,

$\text{d}=\text{a}_2-\text{a}_1=\frac{4\pi}{(\text{p}-\text{q})}-\frac{2\pi}{(\text{p}-\text{q})}=\frac{2\pi}{(\text{p}-\text{q})}$

$\text{d}=\text{a}_3-\text{a}_2=\frac{6\pi}{(\text{p}-\text{q})}-\frac{4\pi}{(\text{p}-\text{q})}=\frac{2\pi}{(\text{p}-\text{q})}$

$\text{d}=\text{a}_4-\text{a}_3=\frac{8\pi}{(\text{p}-\text{q})}-\frac{6\pi}{(\text{p}-\text{q})}=\frac{2\pi}{(\text{p}-\text{q})}$

And so on.

Thus, xx forms a series in AP.

1. $\frac{\pi}{3}$

Solution:

Given: $\cos\text{x}+\sqrt{3}\sin\text{x}=2\ ......(1)$

This equation is of the form $\text{a}\cos\text{x}+\text{b}\sin\text{x}=\text{c},$ where $\text{a}=1,\ \text{b}=\sqrt{3}$ and $\text{c}=2.$

Let:

$\text{a}=\text{r}\ \cos\alpha$ and $\text{b}=\sin\alpha$

Now,

$1=\text{r}\ \cos\alpha,\sqrt{3}=\text{r}\sin\alpha$

$\Rightarrow\text{r}=\sqrt{\text{a}^2+\text{b}^2}=\sqrt{1+3}=\sqrt{4}=2$

And,

$\tan\alpha=\frac{\text{b}}{\text{a}}$

$\Rightarrow\tan\alpha=\frac{\sqrt{3}}{1}$

$\Rightarrow\tan\alpha\sqrt{3}$

$\Rightarrow\alpha=\frac{\pi}{3}$

On putting $\text{a}=1=\text{r}\ \cos\alpha$ and $\text{b}=\sqrt{3}=r\ \sin\alpha$ in equation (1), we get:

$\Rightarrow\text{r}\cos(\text{x}-\alpha)=2$

$\Rightarrow2\cos(\text{x}-\frac{\pi}{3})=2$

$\Rightarrow\cos(\text{x}-\frac{\pi}{3})=1$

$\Rightarrow\cos(\text{x}-\frac{\pi}{3})=\cos0$

$\Rightarrow\text{x}=\frac{\pi}{3}=2\text{n}\pi\pm0$

$\Rightarrow\text{x}=2\text{n}\pi\pm\frac{\pi}{3}$

For $\text{n}=0,\text{x}=\frac{\pi}{3}.$

$\therefore\text{x}=\frac{\pi}{3}$

2. $\text{n}\pi+(-1)^\text{n}\frac{\pi}{6},\text{n}\in\text{Z}$

Solution:

Given equation:

$\cot\text{x}-\tan\text{x}=\sec\text{x}$

$\Rightarrow\frac{\cos\text{x}}{\sin\text{x}}-\frac{\sin\text{x}}{\cos\text{x}}=\frac{1}{\cos\text{x}}$

$\Rightarrow\frac{\cos^2\text{x}-\sin^2}{\sin\text{x}\cos​x​}=\frac{1}{\cos\text{x}}$

$\Rightarrow\cos^2​\text{x}​\sin^2=\sin\text{x}$

$\Rightarrow(1-\sin^2\text{x})-\sin^2\text{x}=\sin\text{x}$

$\Rightarrow1-2\sin^2=\sin\text{x}$

$\Rightarrow2\sin^2\text{x}+\sin\text{x}-1=0$

$\Rightarrow2\sin^2\text{x}+2\sin\text{x}-\sin\text{x}-1=0$

$\Rightarrow2\sin\text{x}(\sin\text{x}+1)-1(\sin\text{x}+1)=0$

$\Rightarrow(\sin\text{x}+1)(2\sin\text{x}-1)=0$

$\Rightarrow\sin\text{x}+1=0$ or $2\sin\text{x}-1=0$

$\Rightarrow\sin\text{x}=-1$ or $\sin\text{x}=\frac{1}{2}$

Now,

$\sin\text{x}=\frac{1}{2}\Rightarrow\sin\text{x}=sin\frac{3\pi}{2}\Rightarrow\text{x}=\text{m}\pi+(-1)^\text{m}\frac{3\pi}{2},\ \text{m}\in\text{Z}$

And

$\sin\text{x}=\frac{1}{2}\Rightarrow\sin\text{x}=\sin\frac{\pi}{6}\Rightarrow\text{x}=\text{n}\pi+(-1)^\text{n}\frac{\pi}{6},\text{n}\in\text{Z}$

$\therefore\text{x}=\text{n}\pi+(-1)^\text{n}\frac{\pi}{6},\text{n}\in\text{Z}$

2. 2

Solution:

$\sin^2\text{x}-\cos\text{x}=\frac{1}{4}$

$\Rightarrow(1-\cos^2\text{x})-\cos\text{x}=\frac{1}{4}$

$\Rightarrow4-4\cos^2\text{x}-4\cos\text{x}=1$

$\Rightarrow4\cos^2-4\cos\text{x}=1$

$\Rightarrow4\cos^2\text{x}+4\cos\text{x}-3=0$

$\Rightarrow4\cos^2\text{x}+6\cos\text{x}-2\cos\text{x}-3=0$

$\Rightarrow2\cos\text{x}(2\cos\text{x}+3)-1(2\cos\text{x}-1)=0$

$\Rightarrow(2\cos\text{x}+3)(2\cos\text{x}-1)=0$

$\Rightarrow2\cos\text{x}+3=0$ or $2\cos\text{x}-1=0$

$\Rightarrow\cos\text{x}=-\frac{3}{2}$  is not possible.

$\therefore\cos\text{x}=\frac{1}{2}$

$\Rightarrow\cos\text{x}=\cos\frac{\pi}{3}$

$\Rightarrow\text{x}=2\text{n}\pi\pm\frac{\pi}{3}$

Now for n = 0 and 1, the values of x are $\frac{\pi}{3},\ \frac{5\pi}{3}$ and $\frac{7\pi}{3},$ but $\frac{7\pi}{3}$  is not in $[0,\ 2\pi]$

Hence, there are two solutions in $[0,\ 2\pi].$

3. 6

Solution:

Given:

$3\sin^2\text{x}-7\sin\text{x}+2=0$

$\Rightarrow3\sin^2\text{x}-6\sin\text{x}-\sin\text{x}+2=0$

$\Rightarrow3\sin\text{x}(\sin\text{x}-2)-1(\sin\text{x}-2)=0$

$\Rightarrow(3\sin\text{x}-1)(\sin\text{x}-2)=0$

$\Rightarrow3\sin\text{x}-1=0$ or $\sin\text{x}-2=0$

Now, $\sin\text{x}=2$ is not possible, as the value of sin xsin x lies between -1 and 1.

$\Rightarrow\sin\text{x}=\frac{1}{4}$

Also, sin x is positive only in first two quadrants. Therefore, sin x is positive twice in the interval $[0,\ \pi].$

Hence, it is positive six times in the interval $[0,\ 5\pi]$ viz $[0,\ \pi],$ $2\pi,\ 3\pi$ and $[4\pi,\ 5\pi].$

3. $\text{x}=\text{n}\pi\frac{\pi}{3},\text{n}\in\text{Z}$

Solution:

Given:

$7\cos^2\text{x}+3\sin^2\text{x}=4$

$\Rightarrow7\cos^2\text{x}+3\Big(1-\cos^2\text{x}\Big)=4$

$\Rightarrow7\cos^2\text{x}+3-3\cos^2\text{x}=4$

$\Rightarrow4\cos^2\text{x}+3=4$

$\Rightarrow4\Big(1-\cos^2\text{x}\Big)=3$

$\Rightarrow4\sin^2\text{x}=3$

$\Rightarrow\sin^2\text{x}=\frac{3}{4}$

$\Rightarrow\sin\text{x}=\frac{\sqrt{3}}{2}$

$\Rightarrow\sin\text{x}=\sin\frac{\pi}{3}$

$\Rightarrow\text{x}=\text{n}\pi\pm\frac{\pi}{3},\text{n}\in\text{Z}$

4. 2

Solution:

Given equation:

$\tan\text{x}+\tan2\text{x}=-\tan3\text{x}+\tan\text{x}\tan2\text{x}\tan3\text{x}$

$\Rightarrow\tan\text{x}+\tan2\text{x}=-\tan3\text{x}+\tan\text{x}\tan2\text{x}\tan3\text{x}$

$\Rightarrow\tan\text{x}+\tan2\text{x}=-\tan3\text{x}\Big(1-\tan\text{x}\tan2\text{x}\Big)$

$\Rightarrow\frac{\tan\text{x}+\tan2\text{x}}{1-\tan\text{x}\tan2\text{x}}=-\tan3\text{x}$

$\Rightarrow\tan(\text{x}+2\text{x})=-\tan3\text{x}$

$\Rightarrow\tan3\text{x}=-\tan3\text{x}$

$\Rightarrow2\tan3\text{x}=0$

$\Rightarrow\tan3\text{x}=0$

$\Rightarrow3\text{x}=\text{n}\pi$

$\Rightarrow\text{x}=\frac{\text{n}\pi}{3}$

Now,

$\text{x}=\frac{\pi}{3},\ \text{n}=1$

$\text{x}=\frac{2\pi}{3},\ \text{n}=2$

$\text{x}=\frac{3\pi}{3}=180^\circ,$ which is not possible, as it is not in the interval $(0,\ 2\pi).$

Hence, the number of solutions of the given equation is 2.

4. None of these

Solution:

Given equation: $\text{e}^{\sin\text{x}}-\text{e}^{-\sin\text{x}}-4=0,$

Let:

$\text{e}^{\sin\text{x}}=\text{y}$

Now,

$\text{y}-\text{y}^{-1}-4=0$

$\Rightarrow\text{y}^2-4\text{y}-1=0$

$\therefore\text{y}=\frac{4\pm\sqrt{16+4}}{2}$

$\Rightarrow\text{y}=\frac{4\pm\sqrt{20}}{2}$

$\Rightarrow\text{y}=\frac{4\pm2\sqrt{5}}{2}=2\pm\sqrt{5}$

And

$\text{y}=\text{e}^{\sin\text{x}}$

$\Rightarrow\text{y}^{\sin\text{x}}={2\pm\sqrt{5}}$

Taking log on both sides, we get:

$\sin\text{x}=\log_\text{e}\big(2\pm\sqrt{5}\big)$

$\Rightarrow\sin\text{x}=\log_{e}\big(2+\sqrt{5}\big)$ or $\sin\text{x}=\log_\text{e}\big(2-\sqrt{5}\big)$

$\Rightarrow\sin\text{x}=\log_{e}\big(4.24\big)$ or $\sin\text{x}=\log_\text{e}\big(-0.24\big)$

$\log(4.24)>1$ and $\sin\text{x}$ cannot be greater than 1.

In the other case, the log of negative term occurs, which is not defined.

4. $\Big(\frac{5\pi}{4},\ \frac{7\pi}{4}\Big)$

Solution:

Given:

$\cos^2\text{x}+\sin\text{x}+1=0$

$\Rightarrow(1-\sin^2\text{x})+\sin\text{x}+1=0$

$\Rightarrow1-\sin^2\text{x}+\sin\text{x}+1=0$

$\Rightarrow\sin^2\text{x}-\sin\text{x}-2=0$

$\Rightarrow\sin^2\text{x}-2\sin\text{x}+\sin\text{x}-2=0$

$\Rightarrow\sin\text{x}(\sin\text{x}-2)+1(\sin\text{x}-2)=0$

$\Rightarrow(\sin\text{x}-2)(\sin\text{x}+1)=0$

$\Rightarrow\sin\text{x}-2=0$ or $\sin\text{x}+1=0$

$\Rightarrow\sin\text{x}=2$ or $\sin\text{x}=-1$

$\sin\text{x}=2$ is not possible.

$\Rightarrow\sin\text{x}=-1$

$\therefore\sin\text{x}=\sin\frac{3\pi}{2}$

$\Rightarrow\text{x}=\text{n}\pi+(-1)^\text{n}\frac{3\pi}{2},\ \text{n}\in\text{Z}$

The values of x lies in the third and fourth quadrants.

Hence, x lies in $\Big(\frac{5\pi}{4},\ \frac{7\pi}{4}\Big).$

1. $\frac{5\pi}{6}$

Solution:

Given:

$2\sin^2\text{x}+\sqrt{3}\cos\text{x}+1=0$

$\Rightarrow2(1-\cos^2\text{x})+\sqrt{3}\cos\text{x}+1=0$

$\Rightarrow2-2\cos^2\text{x}+\sqrt{3}\cos\text{x}+1=0$

$\Rightarrow2\cos^2\text{x}-\sqrt{3}\cos\text{x}-3=0$

$\Rightarrow2\cos^2\text{x}-2\sqrt{3}\cos\text{x}+\sqrt{3}\cos\text{x}-3=0$

$\Rightarrow2\cos\text{x}\Big(\cos\text{x}-\sqrt{3}\Big)+\sqrt{3}\Big(\cos\text{x}-\sqrt{3}\Big)=0$

$\Rightarrow\Big(2\cos\text{x}+\sqrt{3}\Big)\Big(\cos\text{x}-\sqrt{3}\Big)=0$

$\therefore\cos\text{x}+\sqrt{3}=0$ or, $\cos\text{x}=\sqrt{3}$ is not possible.

$\Rightarrow\cos\text{x}=\cos\Big(\frac{\pi}{6}\Big)$

$\Rightarrow\text{x}=2\text{n}\pi\pm\frac{5\pi}{6},\ \text{n}\in\text{Z}$

For n = 0, the value of x is $\pm\frac{5\pi}{6}.$

Hence, the smallest positive angle is $\frac{5\pi}{6}.$

4.  $\Big(\frac{5\pi}{4},\ \frac{7\pi}{4}\Big)$

Solution:

Given equation:

$\cos^2\text{x}+\sin\text{x}+1=0$

$\Rightarrow(1-\sin^2\text{x})+\sin\text{xx}+1=0$

$\Rightarrow2-\sin^2\text{x}+\sin\text{x}=0$

$\Rightarrow\sin^2\text{x}=\sin\text{x}-2=0$

$\Rightarrow\sin^2\text{x}-2\sin\text{x}+\sin\text{x}-2=0$

$\Rightarrow\sin\text{x}(\sin\text{x}-2)+1(\sin\text{x}-2)=0$

$\Rightarrow(\sin\text{x}-2)(\sin\text{x}+1)=0$

$\Rightarrow\sin\text{x}-2=0$ or $\sin\text{x}+1=0$

$\Rightarrow\sin\text{x}=2$ or $\sin\text{x}=-1$

Now, $\sin\text{x}=2$ is not possible.

And,

$\Rightarrow\sin\text{x}=\sin\frac{3\pi}{2}$

$\Rightarrow\text{x}=\text{n}\pi+(-1)^\text{n}\frac{3\pi}{2}$

For $\text{n}=0,\ \text{x}=\frac{3\pi}{2},$ for $\text{n}=1,\ \text{x}=\frac{7\pi}{2}$ and so on.

Hence,  $\frac{3\pi}{2}$ lies in the interval $\Big(\frac{5\pi}{4},\ \frac{7\pi}{4}\Big).$

2. $\text{x}=\frac{2\pi}{3},\ \frac{4\pi}{3}$

Solution:

Given equation:

$\cos\text{x}=-\frac{1}{2}$

$\Rightarrow\cos\text{x}=\cos\frac{2\pi}{3}$

$\Rightarrow\text{x}=\frac{2\pi}{3}$

or

$\cos\text{x}=\cos\frac{4\pi}{3}$

$\Rightarrow\text{x}=\frac{4\pi}{3}$

so, both $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$ line in $0<\text{}\text{x}<2\pi.$

4.  $\frac{\pi}{3}$

Solution:

Given equation:

$\cot\text{x}+\sqrt{3}\sin​​​​\text{x}=2\ .....(1)$

Thus, the equation is of the form $\text{a}\cos\text{x}+\text{b}\sin\text{x}=\text{c},$ where $\text{a}=1,\ \text{b}=\sqrt{3}$ and $\text{c}=3$

Let:

$\text{a}=\text{r}\cos\alpha$ and $\text{b}=\text{r}\sin\alpha$

$1=\text{r}\cos\alpha$ and $\sqrt{3}=\text{r}\sin\alpha$

$\Rightarrow\text{r}\sqrt{\text{a}^2+\text{b}^2}=\sqrt{\Big(\sqrt{3}\Big)^2+1^2}=2$ and

$\tan\alpha=\frac{\text{b}}{\text{a}}\Rightarrow\tan\alpha=\frac{\sqrt{3}}{1}\Rightarrow\tan\alpha\tan\frac{\pi}{3}\Rightarrow\alpha=\frac{\pi}{3}$

On putting $\text{a}=1=\text{r}\cos\alpha$ and $\text{b}=\sqrt{3}=\text{r}\sin\alpha$  in equation (1) we get:

$\text{r}\cos\alpha\cos\text{x}+\text{r}\sin\alpha\sin\text{x}=2$

$\Rightarrow​​\text{r}\cos(\text{x}-\alpha)=2$

$\Rightarrow\text{r}\cos(\text{x}-\frac{\pi}{3})=2$

$\Rightarrow\text{2}\cos(\text{x}-\frac{\pi}{3})=2$

$\Rightarrow\text{}\cos(\text{x}-\frac{\pi}{3})=1$

$\Rightarrow\text{}\cos(\text{x}-\frac{\pi}{3})=\cos0$

$\Rightarrow​\text{x}​-\frac{\pi}{3}=0$

$\Rightarrow\text{x}=\frac{\pi}{3}$ 

4. $\text{n}\pi+(-1)^\text{n}\frac{\pi}{4}-\frac{\pi}{3},\ \text{n}\in\text{Z}$

Solution:

Given equation:

$\sqrt{3}\cos​​​\text{x}​+\sin\text{x}\sqrt{2}\ .....(1)$

Thus, is of the  form $\text{a}\cos\text{x}+\text{b}\sin\text{x}=\text{c},$ where $\text{a}=\sqrt{3},\ \text{b}=1$ and $\text{c}=\sqrt2$

Let:

$\text{a}=\text{r}\sin\alpha$ and $\text{b}=\text{r}\cos\alpha$

Now,

$\text{r}=\sqrt{\text{a}^2+\text{b}^2}=\sqrt{\big(\sqrt{3}\big)^2+1^2}=2$ 

And

$\tan\alpha=\frac{\sqrt{3}}{1}$

$\Rightarrow\tan\alpha=\frac{\sqrt{3}}{1}$

$\Rightarrow\tan\alpha=\tan\frac{\pi}{3}$

$\Rightarrow\alpha=\frac{\pi}{3}$

Putting $\text{a}=\sqrt{3}=\text{r}\sin\alpha$ and  $\text{b}=1=\text{r}\cos\alpha$ in equation (i), we get: 

$\text{r}\cos\text{x}\sin\alpha+\text{r}\sin\text{x}\cos\alpha=\sqrt{2}$

$\Rightarrow\text{r}\sin(\text{x}+\alpha)=\sqrt{2}$

$\Rightarrow2\sin(\text{x}+\alpha)=\sqrt{2}$

$\Rightarrow\sin\big(\text{x}+\frac{\pi}{3}\big)=\frac{1}{\sqrt{2}}$

$\Rightarrow\sin\big(\text{x}+\frac{\pi}{3}\big)=\cos\frac{\pi}{4}$

$\Rightarrow\text{x}\frac{\pi}{3}=\text{n}\pi+(-1)^\text{n}\frac{\pi}{4},\ \text{n}\in\text{Z}$

$\Rightarrow\text{x}​​\text{n}\pi+(-1)^\text{n}\frac{\pi}{4}-\frac{\pi}{3},\ \ \text{n}\in\text{Z}$

3. 6

Solution:

Given:

$\Rightarrow\cos(5\text{x}-2\text{x})\tan5\text{x}=\sin(5\text{x}+2\text{x})$

$\Rightarrow\tan5\text{x}=\frac{\sin(5\text{x}+2\text{x})}{\cos(5\text{x}-2\text{x})}$

$\Rightarrow\tan5\text{x}=\frac{\sin5\text{x}\cos2\text{x}+\cos5\text{x}\sin2\text{x}}{\cos5\text{x}\cos2\text{x}+\sin5\text{x}\cos2\text{x}}$

$\Rightarrow\frac{\sin5\text{x}}{\cos5\text{x}}=\frac{\sin5\text{x}\cos2\text{x}+\cos5\text{x}\sin2\text{x}}{\cos5\text{x}\cos2\text{x}+\sin5\text{x}\cos2\text{x}}$

$\Rightarrow\sin5​\text{x}\cos5\text{x}+\sin^25\text{x}\sin2\text{x}\\=\sin5\text{x}\cos5\text{x}\cos2\text{x}+\cos^25\text{x}\sin2\text{x}$

$\Rightarrow\sin^25\text{x}\sin2\text{x}=\cos^25\text{x}\sin2\text{x}$

$\Rightarrow\Big(\sin^25\text{x}\cos^25\text{x}\Big)\sin2\text{x}=0$

$\Rightarrow\Big(\sin5\text{x}-\cos5\text{x}\Big)\Big(\sin5\text{x}+\cos5\text{x}\Big)\sin2\text{x}=0$

$\Rightarrow\sin5\text{x}-\cos5\text{x}=0,\ \sin5\text{x}+\cos5\text{x}=0$ or $\sin2\text{x}=0$

$\Rightarrow\frac{\sin5\text{x}}{\cos5\text{x}}=1,\ \frac{\sin5\text{x}}{\cos5\text{x}}=-1$ or $\sin2\text{x}=0$

Now,

$\Rightarrow\tan5\text{x}=\tan\frac{\pi}{4}$

$\Rightarrow5\text{x}=\text{n}\pi+\frac{\pi}{4},\ \text{n}\in\text{Z}$

$\Rightarrow\text{x}=\frac{\text{n}\pi}{5}+\frac{\pi}{20},\text{n}\in\text{Z}$

F or n = 0, 1 and 2, the values of x are $\frac{\pi}{20},\ \frac{\pi}{4}$ and $\frac{9\pi}{20},$ respectively. 

Or,

$\tan5\text{x}=1$

$\Rightarrow\tan5\text{x}=\tan\frac{3\pi}{4}$

$\Rightarrow5\pi=\text{n}\pi+\frac{3\pi}{4},\ \text{n}\in\text{Z}$

$\Rightarrow​x\text{x}=\frac{\text{n}\pi}{5}+\frac{3\pi}{20},\ \text{n}\in\text{Z}$

For n = 0 and 1, the values of x are $\frac{3\pi}{20}$ and $\frac{7\pi}{20},$ respectively.

And,

$\sin2\text{x}=\sin0$

$\Rightarrow\sin2\text{x}=\sin0$

$\Rightarrow2\text{x}=\text{n}\pi,\ \text{n}\in\text{Z}$

$\Rightarrow\text{x}=\frac{\text{n}\pi}{2},\ \text{n}\in\text{Z}$

For n = 0, the value of x is 0.

Also, for the odd multiple of $\frac{\pi}{2,}$ tanx is not defined.

Hence, there are six solutions.