Question 14 Marks
Solve the following equations: $\cos\text{x}+\sin\text{x}=\cos2\text{x}+\sin2\text{x}$
Answer$\cos\text{x}+\sin\text{x}=\cos2\text{x}+\sin2\text{x}$$\Rightarrow\cos2\text{x}-\cos\text{x}+\sin2\text{x}-\sin\text{x}=0$
$\Rightarrow-2\sin\frac{3\text{x}}{2}\sin\frac{x}{2}+2\cos\frac{3\text{x}}{2}\sin\frac{\text{x}}{2}=0$
$\Rightarrow2\sin\frac{\pi}{\text{x}}\Big(\cos\frac{3\pi}{2}-\sin\frac{3\pi}{2}\Big)=0$
$\Rightarrow2\sin\frac{\text{x}}{2}=0$ or $\cos\frac{3\text{x}}{2}-\sin\frac{3\text{x}}{2}=0$
$\Rightarrow\sin\frac{\text{x}}{2}=0$ or $\cos\frac{3\text{x}}{2}=\sin\frac{3\text{x}}{2}$
$\Rightarrow\frac{\text{x}}{2}=\text{n}\pi$ or $\tan\frac{3\text{x}}{2}=1$
$\Rightarrow\text{x}=2\text{n}\pi$ or $\tan\frac{3\text{x}}{2}=\tan\frac{\pi}{4}$
$\Rightarrow\text{x}=\text{n}\pi$ or $\frac{3\pi}{2}-\text{n}\pi+\frac{\pi}{4}$
$\Rightarrow\text{x}=2\text{n}\pi$ or $3\text{x}=2\text{n}\pi+\frac{\pi}{2}$
$\Rightarrow\text{x}=2\text{n}\pi$ or $\text{x}=\frac{2\text{n}\pi}{3}+\frac{\pi}{6},\text{n}\in\text{Z}$
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Solve the following equations: $\sin\text{x}\ \tan\text{x}-1\tan\text{x}-\sin\text{x}$
Answer$\sin\text{x}\ \tan\text{x}-1=\tan\text{x}-\sin\text{x}$ $=\sin\text{x}\ \text{tan}\text{x}-\tan\text{x}+\sin\text{x}-1=0$ $\Rightarrow\tan\text{x}(\sin\text{x}-1)+1(\sin\text{x})-1=0$ $\Rightarrow(\tan\text{x}+1)(\sin\text{x}-1)=0$ $\Rightarrow(\tan\text{x}+1)=0$ or $(\sin\text{x}-1)=0$ $\Rightarrow\tan\text{x}=-1$ or $\sin\text{x}=1$ $\Rightarrow\tan\text{x}=\tan\frac{3\pi}{4}$ or $\sin\text{x}=\sin\frac{\pi}{2}$ $\Rightarrow\text{x}=\text{n}\pi+\frac{3\pi}{4}$ or $\text{x}=\text{n}\pi+(-1)^\text{n}\frac{\pi}{2},\ \text{n}\in\text{Z}$
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Solve the following equations: $3\sin2\text{x}-5 \sin\text{x}\cos \text{x} + 8 \cos2\text{x = 2}$
Answer$3\sin^2\text{x}-5\sin\text{x}\cos\text{x}+8\cos^2\text{x}=2$ $\Rightarrow3\sin^2\text{x}-5\sin\text{x}\cos\text{x}+3\cos^\text{x}2+5\cos^2\text{x}-2=0$ $\Rightarrow3(\sin^2\text{x}\cos^2\text{x})-5\sin\text{x}\cos\text{x}+5\cos^2\text{x}-2=0$ $\Rightarrow3-5\sin\text{x}\cos\text{x}+5\cos^2\text{x}-2=0$ $\Rightarrow5\cos^2\text{x}-5\sin\text{x}\cos\text{x}+1=0$ $\Rightarrow5(1-\sin^2\text{x})-5\sin\text{x}\cos\text{x}+1=0$ $\Rightarrow5-5\sin^2\text{x}-5\sin\text{x}\cos\text{x}+1=0$ $\Rightarrow5\sin^2\text{x}+5\sin\text{x}\cos\text{x}-6=0$ Dividing by $\cos^2\text{x},\ $ we get $\Rightarrow5\tan^2\text{x}+5\tan\text{x}-6\sec^2\text{x}=0$ $\Rightarrow5\tan^2\text{x}+5\tan\text{x}-6-6\tan^2\text{x}=0$ $\Rightarrow-\tan^2\text{x}+5\tan\text{x}-6=0$ $\Rightarrow\tan^2\text{x}-5\tan\text{x}+6=0$ $\Rightarrow\tan^2\text{x}-3\tan\text{x}-2\tan\text{x}+6=0$ $\Rightarrow(\tan\text{x}-3)=0$ or $\tan\text{x}=2$ $\Rightarrow\text{x}=\text{n}\pi+\tan^{-1}3$ or $\text{x}=\text{n}\pi+\tan6{-1}2,\ \text{n}\in\text{Z}$
View full question & answer→Question 44 Marks
Solve the following equations: $\sin\text{x}-3\sin2\text{x}+\sin3\text{x}=\cos\text{x}-3\cos2\text{x}+\cos3\text{x}$
Answer$\sin\text{x}-3\sin2\text{x}+\sin3\text{x}=\cos\text{x}-3\cos2\text{x}+\cos3\text{x}$ $(\sin\text{x}+\sin3\text{x})-3\sin2\text{x}-(\cos\text{x}+\cos3\text{x})+3\cos2\text{x}=0$ $2\sin2\text{x}\cos\text{x}-3\sin2\text{x}-2\cos2\text{x}\cos\text{x}+3\cos2\text{x}=0$ $\sin2\text{x}(2\cos\text{x}-3)-\cos2\text{x}(2\cos\text{x}-3)=0$ $(2\cos\text{x}-3)(\sin2\text{x}-\cos2\text{x})=0$ $\cos\text{x}=\frac{3}{2}$ or $\sin2\text{x}-\cos2\text{x}-\cos2\text{x}=0$ but $\cos\text{x}\in[-11]\Rightarrow\cos\text{x}\not=\frac{3}{2}$ $\sin2\text{x}=\cos2\text{x}$ $2\text{x}=\text{n}\pi+\frac{\pi}{4}$ $\text{x}=\frac{\text{n}\pi}{2}+\frac{\pi}{8}$
View full question & answer→Question 54 Marks
Solve the following equations: $4\sin\text{x}\cos\text{x}+2\sin\text{x}+2\cos\text{x}+1=0$
Answer$4\sin\text{x}\cos\text{x}+2\sin\text{x}+2\cos\text{x}+1=0$ $\Rightarrow2\sin\text{x}(2\cos\text{x}+1)+1(2\cos\text{x}+1)=0$ $\Rightarrow(2\sin\text{x}+1)(2\cos\text{x+1})=0$ $\Rightarrow2\sin\text{x}+1=0$ or $2\cos\text{x}=-\frac{1}{2}$ $\Rightarrow\sin\text{x}=\sin\frac{7\pi}{6}$ or $\cos\text{x}=\frac{2\pi}{3}$ $\Rightarrow\text{x}=\text{n}\pi+(-1)^\text{n}\frac{7\pi}{6}$ or $\text{x}=2\text{n}\pi\pm\frac{2\pi}{3},\text{n}\in\text{Z}$
View full question & answer→Question 64 Marks
Solve the following equations: $2^{\sin^2\text{x}}+2\cos^{2\text{x}}=2\sqrt{2}$
Answer$2^{\sin^2\text{x}}+2\cos^{2\text{x}}=2\sqrt{2}$ $\Rightarrow2^{\sin^2\text{x}}+2\cos^{-1\sin^2\text{x}}=2\sqrt{2}$ $\Rightarrow2^{\sin2\text{x}}+\frac{2}{2\sin^2\text{x}}2\sqrt{2}$ $\text{Let}2^{\sin^2\text{x}}+\frac{2}{2^{\sin2\text{x}}}=\text{y}$ $\Rightarrow\text{y}+\frac{2}{\text{y}}=2\sqrt{2}$ $\Rightarrow\text{y}^2+2=2\sqrt{2\text{y}}$ $\Rightarrow\text{y}^2-2\sqrt{2\text{y}}+2=0$ $\Rightarrow\text{y}^2-2\sqrt{2\text{y}}-\sqrt{2}\text{y}-\sqrt{2\text{y}}+2=0$ $\Rightarrow\text{y}\Big(\text{y}-\sqrt{2}\Big)-\sqrt{2}\Big(\text{y}-\sqrt{2}\Big)=0$ $\Rightarrow\Big(\text{y}-\sqrt{2}\Big)^2=0$ $\Rightarrow\Big(\text{y}-\sqrt{2}\Big)=0$ $\Rightarrow\text{y}=\sqrt{2}$ $\Rightarrow2^{\sin^2\text{x}}=2\frac{1}{2}$ $\Rightarrow\sin^2\text{x}=\frac{1}{2}$ $\Rightarrow\sin^2\text{x}=\sin^2\frac{\pi}{4}$ $\Rightarrow\text{x}=\text{n}\pi\pm\frac{\pi}{4},\text{n}\in\text{Z}$
View full question & answer→Question 74 Marks
Solve the following equations: $3\tan\text{x}+\cot\text{x}=5\ \text{cosec }\text{x}$
Answer$3\tan\text{x}+\cot\text{x}=5\ \text{cosec}\ \text{x}$ $\Rightarrow\frac{3\sin\text{x}}{\cos\text{x}}+\frac{\cos\text{x}}{\sin\text{x}}=\frac{5}{\sin\text{x}}$ $\Rightarrow\frac{3\sin^2+\cos^2\text{x}}{\cos\text{x}\sin\text{x}}=\frac{5}{\sin\text{x}}$ $\Rightarrow3(1-\cos^2\text{x})+\cos^2\text{x}=5\cos\text{x}$ $\Rightarrow3-3\cos^2\text{x}+\cos^2\text{x}=5\cos\text{x}$ $\Rightarrow2\cos^2\text{x}+5\cos\text{x}-3=0$ $\Rightarrow2\cos^2\text{x}+6\cos\text{x}-\cos\text{x}-3=0$ $\Rightarrow2\cos\text{x}(\cos\text{x}+3)-1(\cos\text{x}+3)=0$ $\Rightarrow(2\cos\text{x}-1)(\cos\text{x}+3)=0$ $\Rightarrow(2\cos\text{x}-1)=0$ or $\cos\text{x}+3=0$ $\Rightarrow\cos\text{x}=\frac{1}{2}$ or $\cos\text{x}=-3$$\cos\text{x}=-3$ is not possible $(\therefore-1\leq\cos\text{x}\leq1)$
$\Rightarrow\cos\text{x}=\cos\frac{\pi}{3}$
$\Rightarrow\text{x}=2\text{n}\pi\pm\frac{\pi}{3},\ \text{n}\in\text{Z}$
View full question & answer→Question 84 Marks
Solve the following equations: $3-2\cos\text{x}-4\sin\text{x}-\cos2\text{x}+\sin2\text{x}=0$
Answer$3-2\cos\text{x}-4\sin\text{x}-\cos2\text{x}+\sin2\text{x}=0$ $\Rightarrow3-2\cos\text{x}-4\sin\text{x}-(1-2\sin^2\text{x})+2\sin\text{x}\cos\text{x}=0$ $\Rightarrow3-2\cos\text{x}-4\sin\text{x}-1+2\sin^2\text{x}+2\sin\text{x}\cos\text{x}=0$ $\Rightarrow(2\sin^2\text{x}-4\sin\text{x}+2)+2\cos\text{x}(\sin\text{x}-1)=0$ $\Rightarrow2(\sin^2\text{x}-2\sin\text{x}+1)+2\cos\text{x}(\sin\text{x}-1)=0$ $\Rightarrow2(\sin\text{x}-1)^2+2\cos\text{x}(\sin\text{x}-1)=0$ $\Rightarrow(\sin\text{x}-1)(2\sin\text{x}-2+2\cos\text{x})=0$ $\Rightarrow2(\sin\text{x}-1)(\sin\text{x}+\cos\text{x}-1)=0$ $\Rightarrow(\sin\text{x}-1)=0$ or $(\sin\text{x})+\cos\text{x}-1=0$ $\Rightarrow\sin\text{x}=1$ or $\sin\text{x}+\cos\text{x}=1$ $\Rightarrow\sin\text{x}=\sin\frac{\pi}{2}$ or $\frac{1}{\sqrt{2}}\sin\text{x}+\frac{1}{\sqrt{2}}\cos\text{x}=\frac{1}{\sqrt{2}}$ $\Rightarrow\sin\text{x}=\sin\frac{\pi}{2}$ or $\sin\frac{\pi}{4}\sin\text{x}+\cos\frac{\pi}{4}\cos\text{x}=\cos\frac{\pi}{4}$ $\Rightarrow\sin\text{x}=\sin\frac{\pi}{2}$ or $\cos\Big(\text{x}-\frac{\pi}{4}\Big)=\cos\frac{\pi}{4}$ $\Rightarrow\text{x}=\text{n}\pi+(-1)^\text{n}\frac{\pi}{2}$ or $\text{x}-\frac{\pi}{4}=2\text{n}\pi\pm\frac{\pi}{4},\ \text{n}\in\text{Z}$ $\Rightarrow\text{x}=\text{n}\pi+(-1)^\text{n}$ or $\text{x}=2\text{n}\pi+\frac{\pi}{2}$ or $\text{x}=2\text{n}\pi,\ 2\text{n}\pi,\ \text{n}\in\text{Z}$
View full question & answer→Question 94 Marks
Find the general solutions of the following equations: $\cos4\text{x}=\cos2\text{x}$
AnswerWe have, $\cos4\text{x}=\cos2\text{x}$ $\Rightarrow\cos4\text{x}=\cos2\text{x}=0$ $\Rightarrow2\sin\text{x}.\sin3\text{x}=0$ $\Rightarrow\text{Either}$ $\sin\text{x}=0$ or $\sin3\text{x}=0$ $\Rightarrow\text{x}=\text{n}\pi,\text{n}\in\text{z}$ or $3\text{x}=\text{m}\pi,\text{m}\in$ Thus, $\text{x}=\text{n}\pi$ or $\text{m}\frac{\pi}{3},\text{n,m}\in\text{z}$
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Find the general solutions of the following equations: $3\cos^{2}\text{x}-2\sqrt{3}\sin\text{x}\cos\text{x}-3\sin^{2}\text{x}=0$
AnswerWe have, $3\cos^{2}\text{x}-2\sqrt{3}\sin\text{x}\cos\text{x}-3\sin^{2}\text{x}=0$ $\sqrt{3}\cos^{2}\text{x}-2\sin\text{x}\cos\text{x}-\sqrt{3}\sin^{2}\text{x}=0$ $(\text{Divided by}\sqrt{3})$ $\sqrt{3}\cos^{2}\text{x}-\sin\text{x}\cos\text{x}-3\sin\text{x}\cos\text{x}\sqrt{3}\sin^{2}\text{x}=0$ $\cos\text{x}(\sqrt{3}\cos\text{x}+\sin\text{x})-\sqrt{3}\sin\text{x}(\sqrt{3}\cos\text{x}+\sin\text{x})=0$ $(\sqrt{3}\cos\text{x}+\sin\text{x})(\cos\text{x}-\sqrt{3}\sin\text{x})=0$ $\sqrt{3}\cos\text{x}+\sin\text{x}=0$ or $\cos\text{x}-\sqrt{3}\sin\text{x}=0$ $\tan\text{x}=-\sqrt{3}=-\tan\frac{\pi}{3}$ or $\tan\text{x}=\frac{1}{\sqrt{3}}=\tan\frac{\pi}{6}$ $\text{x}=\text{n}\pi-\frac{\pi}{3}$ or $\text{x}=\text{m}\pi-\frac{\pi}{6}$ $\text{n,m}\in\text{z}$
View full question & answer→Question 114 Marks
Solve the following equations: $\sin2\text{x}-\sin4\text{x}+\sin6\text{x}=0$
Answer$\sin2\text{x}-\sin4\text{x}+\sin6\text{x}=0$ $(\sin2\text{x}+\sin6\text{x})-\sin4\text{x}=0$ $2.\sin\Big(\frac{8\text{x}}{2}\Big).\cos\Big(\frac{4\text{x}}{2}\Big)-\sin4\text{x}=0$ $2\sin4\text{x}.\cos2\text{x}-\sin4\text{x}=0$ $\sin4\text{x}(2\cos2\text{x}-1)=0$ $\sin4\text{x}=0$ or $2\cos2\text{x}-1=0$ $4\text{x}=\text{n}(\pi)$ or $\cos2\text{x}=\frac{1}{2}$ $\text{x}=[\frac{\text{n}\pi}{4}]$ or $\cos2\text{x}=\cos[\frac{\pi}{3}]$ $\text{x}=[\frac{\text{n}\pi}{4}]$ or $\text{x}=\text{n}(\pi)\pm[\frac{\pi}{6}]$
View full question & answer→Question 124 Marks
Solve the following equation: $2\cos^{2}\text{x}-5\cos\text{x}+2=0$
AnswerWe have, $2\cos^{2}\text{x}-5\cos\text{x}+2=0$ $\Rightarrow2\cos^{2}\text{x}-4\cos\text{x}-\cos\text{x}+2=0$ [Use factorization] $\Rightarrow2\cos\text{x}(\cos\text{x}-2)-1(\cos\text{x}-2)=0$ $\Rightarrow(2\cos\text{x}-1)(\cos\text{x}-2)=0$ $\Rightarrow\text{Either}$ $2\cos\text{x}-1=0$ or $\cos\text{x}-2=0$ $\Rightarrow\cos\text{x}=\frac{1}{2}$ or $\cos\text{x}=2$ $\Rightarrow\cos\text{x}=\cos\frac{\pi}{3}$ $\big[$This is not possible as $-1<\cos\text{x}<1\big]$ $\Rightarrow\text{x}=2\text{n}\pi\pm\frac{\pi}{3},\text{n}\in\text{z}$ Thus, $\text{x}=2\text{n}\pi\pm\frac{\pi}{3},\text{n}\in\text{z}$
View full question & answer→Question 134 Marks
Solve the following equations: $\cot\text{x}+\tan\text{x}=2$
Answer$\cot\text{x}+\tan\text{x}=2$ $\Rightarrow\frac{1}{\tan\text{x}}+\tan\text{x}=2$ $\Rightarrow\tan^{2}\text{x}+1=2\tan\text{x}$ $\Rightarrow\tan^{2}\text{x}-2\tan\text{x}+1=0$ $\Rightarrow(\tan\text{x}-1)^{2}=0$ $\Rightarrow\tan\text{x}=1=\tan\frac{\pi}{4}$ $\Rightarrow\text{x}=\text{n}\pi+\frac{\pi}{4},\text{n}\in\text{Z}$ $(\tan\text{x}=\tan\alpha\Rightarrow\text{x}=\text{n}\pi+\alpha,\text{n}\in\text{z)}$
View full question & answer→Question 144 Marks
Solve the following equations: $\sin\text{x}+\cos\text{x}=1$
AnswerWe have, $\sin\text{x}+\cos\text{x}=1$ divide both side by $\sqrt{2},$ we get, $\Rightarrow\frac{1}{\sqrt{2}}\sin\text{x}+\frac{1}{\sqrt{2}}\cos\text{x}=\frac{1}{\sqrt{2}}$ $\Rightarrow\sin\frac{\pi}{4}\sin\text{x}+\cos\frac{\pi}{4}\cos\text{x}=\frac{1}{\sqrt{2}}$ $\Rightarrow\cos\Big(\text{x}-\frac{\pi}{4}\Big)=\cos\frac{\pi}{4}$ $\Rightarrow\text{x}=\frac{\pi}{4}=2\text{n}\pi\pm\frac{\pi}{4},\text{n}\in\text{z}$ $\Rightarrow\text{x}=2\text{n}\pi+\frac{\pi}{2}$ or $2\text{n}\pi,\text{n}\in\text{z}$
View full question & answer→Question 154 Marks
Find the general solutions of the following equations: $\sin3\text{x}+\cos2\text{x}=0$
Answer$\cos(2\text{x})=-\sin(3\text{x})$$=-\cos(\frac{\pi}{2}-3\text{x})$
$=\cos(\frac{\pi}{2}+3\text{x})$
$\Rightarrow2\text{n}\pi+2\text{x}=\frac{\pi}{2}+3\text{x}$
$\text{x}=(4\text{m}-1)\frac{\pi}{2},\text{m}\in\text{z}$
or
$\Rightarrow2\text{n}\pi-2\text{x}=\frac{\pi}{2}+3\text{x}$
$\text{x}=(4\text{n}-1)\frac{\pi}{10},\text{n}\in\text{z}$
View full question & answer→Question 164 Marks
Solve the following equations: $\sin\text{x}+\sin2\text{x}+\sin3=0$
AnswerWe have, $\sin\text{x}+\sin2\text{x}+\sin3=0$ $\Rightarrow\sin2\text{x}+2\sin2\text{x}.\cos\text{x}=0$ $\Rightarrow\sin2\text{x}+(1+2\cos\text{x})=0$ $\Rightarrow\text{Either}$ $\sin2\text{x}=0$ or $1+2\cos\text{x}=0$ $\Rightarrow2\text{x}=\text{n}\pi,\text{n}\in\text{z}$ or $\cos\text{x}=-\frac{1}{2}=\cos\Big(\pi-\frac{\pi}{3}\Big)$ $\Rightarrow\text{x}=\frac{\text{n}\pi}{2},\text{n}\in\text{z}$ or $\text{x}=2\text{m}\pi\pm\frac{2\pi}{3},\text{m}\in\text{z}$ Thus, $\text{x}=\frac{\text{n}\pi}{2},\text{n}\in\text{z}$ or $\text{x}=2\text{m}\pi\pm\frac{2\pi}{3},\text{m}\in\text{z} $
View full question & answer→Question 174 Marks
Solve the following equations: $\cos\text{x}\cos2\text{x}\cos3\text{x}=\frac{1}{4}$
AnswerWe have, $\cos\text{x}\cos2\text{x}\cos3\text{x}=\frac{1}{4}$
View full question & answer→Question 184 Marks
Solve the following equations: $\tan\text{x}+\tan2\text{x}=\tan3\text{x}$
Answer$\tan\text{x}+\tan2\text{x}=\tan(\text{x}+2\text{x})$ $\tan\text{x}+\tan2\text{x}-\frac{\tan\text{x}+\tan2\text{x}}{1-\tan\text{x}\tan2\text{x}}=0$ $[\tan\text{x}+\tan2\text{x}]\Big[1-\frac{1}{1-\tan\text{x}\tan2\text{x}}\Big]=0$ $[\tan\text{x}+\tan2\text{x}]\Big[\frac{1-\tan\text{x}\tan2\text{x}-1}{1-\tan\text{x}\tan2\text{x}}\Big]=0$ $[\tan\text{x}+\tan2\text{x}]\Big[\frac{-\tan\text{x}\tan2\text{x}}{1-\tan\text{x}\tan2\text{x}}\Big]=0$ $\tan\text{x}=0$ or $\tan2\text{x}=0$ or $\tan\text{x}+\tan2\text{x}=0$ $\text{x}=\text{n}\pi$ or $\frac{\text{n}\pi}{2}$ or $\tan\text{x}\Big[\frac{1-\tan^{2}\text{x}+2}{1-\tan^{2}\text{x}}\Big]=0$ $\text{x}=\text{n}\pi$ or $\frac{\text{n}\pi}{2}$ or $\tan\text{x}=\pm\sqrt{3}$$\text{x}=\text{m}\pi$ or $\frac{\text{n}\pi}{3}\text{m,n}\in\text{Z}$
View full question & answer→Question 194 Marks
Solve the following equations: $\sin\text{x}+\sin2\text{x}+\sin3\text{x}+\sin4\text{x}=0$
AnswerGiven, $\sin\text{x}+\sin2\text{x}+\sin3\text{x}+\sin4\text{x}=0$$(\sin4\text{x}+\sin2\text{x})+(\sin3\text{x}+\sin\text{x})=0$
Using, $(\sin\text{A}+\sin\text{B})\text{Formula}\Rightarrow$
$2\sin\Big[\frac{(4\text{x}+2\text{x})}{2}\Big]\cos\Big[\frac{4\text{x}-2\text{x}}{2}\Big]+2\sin\Big[\frac{( 3\text{x}+\text{x})}{2}\Big]\cos\Big[\frac{( 3\text{x}-\text{x})}{2}\Big]=0$
$2\sin3\text{x}\cos\text{x}+2\sin2\text{x}\cos\text{x}=0$
$2\cos\text{x}(\sin3\text{x}+\sin2\text{x})=0$
$2\cos\text{x}(2\sin)\Big[\frac{(3\text{x}+2\text{x})}{2}\Big]\cos\Big[\frac{(3\text{x}-2\text{x})}{2}\Big]=0$
$4\cos\text{x}\sin\frac{5\text{x}}{2}\cos\frac{\text{x}}{2}=0$
$\cos\text{x}=0;\sin\frac{5\text{x}}{2}=0;\cos\frac{\text{x}}{2}=0$
$\text{x}=\frac{(2\text{n}+1)\pi}{2};\frac{5\text{x}}{2}=\text{m}\pi;\frac{\text{x}}{2}=\frac{(2\text{x}+1)\pi}{2}$
$\text{x}=\frac{(2\text{n}+1)\pi}{2};\text{x}=\frac{2\text{m}\pi}{5};\text{x}=(2\text{x}+1)\pi,\text{m},\text{r},\text{n}\in\text{Z}$
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Solve the following equations: $2\sin^{2}\text{x}=3\cos\text{x},0\leq\text{x}\leq2\pi$
Answer$2\sin^{2}\text{x}=3\cos\text{x}$$\Rightarrow2(1-\cos^{2}\text{x})=3\cos\text{x}$
$\Rightarrow2\cos^{2}\text{x}+3\cos\text{x}-2=0$
$\Rightarrow(2\cos\text{x}-1)(\cos\text{x}+2)=0$
$\Rightarrow\cos\text{x}=\frac{1}{2}$ or $\cos\text{x}=-2$
But, $\cos\text{x}=-2$ is not possible. $(-1\leq\cos\text{x}\leq1)$
$\therefore\cos\text{x}=\frac{1}{2}=\cos\frac{\pi}{3}$
$\Rightarrow\text{x}=2\text{n}\pi\pm\frac{\pi}{3},\text{n}\in\text{z}$
Putting n=0 and n=1, We get
$\text{x}=\frac{\pi}{3},\frac{5\text{x}}{3}$ $(0\leq\text{x}\leq2\pi)$
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Solve the following equations: $\sqrt{3}\cos\text{x}+\sin\text{x}=1$
AnswerWe have, $\sqrt{3}\cos\text{x}+\sin\text{x}=1$ Divide both side by 2, we get $\frac{\sqrt{3}}{2}\cos\text{x}+\frac{1}{2}\sin\text{x}=\frac{1}{2}$ $\Rightarrow\cos\frac{\pi}{6}\cos\text{x}+\sin\frac{\pi}{6}\sin\text{x}=\frac{1}{2}$ $\Big[\because\sin\frac{\pi}{6}=\frac{1}{2},\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}\Big]$ $\Rightarrow\cos\Big(\text{x}-\frac{\pi}{6}\Big)=\cos\frac{\pi}{3}$ $\Rightarrow\text{x}=\frac{\pi}{6}=2\text{n}\pm\frac{\pi}{3},\text{n}\in\text{z}$ $\Rightarrow\text{x}=2\text{n}\pi\pm\frac{\pi}{3}+\frac{\pi}{6},\text{n}\in\text{z}$$\Rightarrow\text{x}=(4\text{n}+1)\frac{\pi}{2}$ or $(12\text{m}-1)\frac{\pi}{6},\text{n},\text{m}\in\text{z}$
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Solve the following equation: $\sin^{2}\text{x}-\cos\text{x}=\frac{1}{4}$
AnswerWe have, $\sin^{2}\text{x}-\cos\text{x}=\frac{1}{4}$ $\Rightarrow1-\cos^{2}\text{x}-\cos\text{x}=\frac{1}{4}$ $[\because\sin^{2}\text{x}=1-\cos^{2}\text{x}]$ $\Rightarrow\cos^{2}\text{x}+\cos\text{x}-\frac{3}{4}=0$ $\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$ [factorize it] $\Rightarrow2\cos\text{x}(2\cos\text{x}+3)-1(\cos\text{x}+3)=0$ $\Rightarrow(2\cos\text{x}-1)(2\cos\text{x}+3)=0$ $\Rightarrow\text{ Either}$ $2\cos\text{x}-1=0$ or $2\cos\text{x}+3=0$ $\Rightarrow\cos\text{x}=\frac{1}{2}$ or $\cos\text{x}=-\frac{3}{2}$ [This is not possible as $-1<\cos\text{x}<1$] $\Rightarrow\cos\text{x}=\cos\frac{\pi}{3}$ $\Rightarrow\text{x}=2\text{n}\pi\pm\frac{\pi}{3},\text{n}\in\text{z}$
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Solve the following equations: $\tan\text{x}+\tan2\text{x}+\tan3\text{x}=0$
Answer$\tan\text{x}+\tan2\text{x}+\frac{(\tan\text{x}+\tan2\text{x})}{1-\tan\text{x}.\tan2\text{x}}=0$ $[\tan\text{x}+\tan2\text{x}]\Big[1+\frac{1}{1-\tan\text{x}.\tan2\text{x}}\Big]=0$ $\tan\text{x}+\tan2\text{x}(2-\tan\text{x}.\tan2\text{x})=0$ $\tan\text{x}=\tan(-2\text{x})$ or $\tan\text{x}.\tan2\text{x}=0$ $\text{x}=\text{n}\pi-2\text{x}$ or $\tan\text{x}.\frac{2\tan\text{x}}{1-\tan^{2}\text{x}}=2$ $3\text{x}=\text{n}\pi$ or $\frac{2\tan^{2}\text{x}}{1-\tan^{2}\text{x}}=2$ $3\text{x}=\text{n}\pi$ or $2\tan^{2}\text{x}=2-2\tan^{2}\text{x}$ $3\text{x}=\text{n}\pi$ or $4\tan^{2}\text{x}=2$ $\text{x}=\frac{\text{n}\pi}{3}$ or $\tan^{2}\text{x}=\frac{1}{2}$ $\text{x}=\frac{\text{n}\pi}{3}$ or $\text{x}=\text{m}\pi\pm\tan^{-1}(\frac{1}{\sqrt{2}}),\text{n,m}\in\text{Z}$
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Solve the following equations: $\sin\text{x}+\cos\text{x}=\sqrt{2}$
AnswerWe have, $\sin\text{x}+\cos\text{x}=\sqrt{2}$ $\Rightarrow\frac{1}{\sqrt{2}}\sin\text{x}+\frac{1}{\sqrt{2}}\cos\text{x}=1$ $\Rightarrow\sin\frac{\pi}{4}\sin\text{x}+\cos\frac{\pi}{4}\cos\text{x}=1$$\Big[\because\cos\frac{\pi}{4}=\sin\frac{\pi}{4}=\frac{1}{\sqrt{3}}\Big]$ $\Rightarrow\cos\Big(\text{x}-\frac{\pi}{4}\Big)=\cos0^\circ$ $\Rightarrow\text{x}-\frac{\pi}{4}=2\text{n}\pi,\text{n}\in\text{z}$ $\Rightarrow\text{x}=2\text{n}\pi+\frac{\pi}{4},\text{n}\in\text{z}$ $\therefore\text{x}=(8\text{n}+1)\frac{\pi}{4},\text{n}\in\text{z}$
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