Solve the Following Question.(5 Marks)

Question 15 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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Question 25 Marks

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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Question 35 Marks

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 $\text{x}=\text{n}\pi+\tan6{-1}2,\ \text{n}\in\text{Z}$

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Question 45 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 $\cos\text{x}\in[-11]\Rightarrow\cos\text{x}\not=\frac{3}{2}$
$\sin2\text{x}=\cos2\text{x}$
$2\text{x}=\text{n}\pi+\frac{\pi}{4}$
$\text{x}=\frac{\text{n}\pi}{2}+\frac{\pi}{8}$

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Question 55 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 $\text{x}=2\text{n}\pi\pm\frac{2\pi}{3},\text{n}\in\text{Z}$

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Question 65 Marks

Solve the following equations:
$2^{\sin^2\text{x}}+2\cos^{2\text{x}}=2\sqrt{2}$

Answer

$2^{\sin^2\text{x}}+2\cos^{2\text{x}}=2\sqrt{2}$
$\Rightarrow2^{\sin^2\text{x}}+2\cos^{-1\sin^2\text{x}}=2\sqrt{2}$
$\Rightarrow2^{\sin2\text{x}}+\frac{2}{2\sin^2\text{x}}2\sqrt{2}$
$\text{Let}2^{\sin^2\text{x}}+\frac{2}{2^{\sin2\text{x}}}=\text{y}$
$\Rightarrow\text{y}+\frac{2}{\text{y}}=2\sqrt{2}$
$\Rightarrow\text{y}^2+2=2\sqrt{2\text{y}}$
$\Rightarrow\text{y}^2-2\sqrt{2\text{y}}+2=0$
$\Rightarrow\text{y}^2-2\sqrt{2\text{y}}-\sqrt{2}\text{y}-\sqrt{2\text{y}}+2=0$
$\Rightarrow\text{y}\Big(\text{y}-\sqrt{2}\Big)-\sqrt{2}\Big(\text{y}-\sqrt{2}\Big)=0$
$\Rightarrow\Big(\text{y}-\sqrt{2}\Big)^2=0$
$\Rightarrow\Big(\text{y}-\sqrt{2}\Big)=0$
$\Rightarrow\text{y}=\sqrt{2}$
$\Rightarrow2^{\sin^2\text{x}}=2\frac{1}{2}$
$\Rightarrow\sin^2\text{x}=\frac{1}{2}$
$\Rightarrow\sin^2\text{x}=\sin^2\frac{\pi}{4}$
$\Rightarrow\text{x}=\text{n}\pi\pm\frac{\pi}{4},\text{n}\in\text{Z}$

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Question 75 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 $(\therefore-1\leq\cos\text{x}\leq1)$
$\Rightarrow\cos\text{x}=\cos\frac{\pi}{3}$
$\Rightarrow\text{x}=2\text{n}\pi\pm\frac{\pi}{3},\ \text{n}\in\text{Z}$

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Question 85 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 $\text{x}=2\text{n}\pi,\ 2\text{n}\pi,\ \text{n}\in\text{Z}$

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Maths Solve the Following Question.(5 Marks)

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