SATHEE: Trigonometrical Equations 1 Question 30

Trigonometrical Equations 1 Question 30

30. Solve $2 (\cos x + \cos 2 x) + (1 + 2 \cos x) \sin 2 x$

$= 2 \sin x, - π \leq x \leq π$

(1978, 3M)

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Answer:

Correct Answer: 30. $x = - π, - \frac{π}{2}, - \frac{π}{3}, \frac{π}{3}, π$

Solution:

Given that,

$2 \cos x + 2 \cos 2 x + \sin 2 x + \sin 3 x + \sin x - 2 \sin x = 0$

$∴ 2 \cos x + 2 \cos 2 x + 2 \sin x \cos x + (\sin 3 x - \sin x) = 0$

$\Rightarrow 2 \cos x + 2 \cos 2 x + 2 \sin x \cos x + 2 \cos 2 x \sin x = 0$

$\Rightarrow 2 \cos x (1 + \sin x) + 2 \cos 2 x (1 + \sin x) = 0$

$\Rightarrow 2 (1 + \sin x) (\cos x + \cos 2 x) = 0$

$\Rightarrow 4 (1 + \sin x) \cos \frac{3 x}{2} \cos \frac{x}{2} = 0$

$∴ 1 + \sin x = 0$

or $\cos \frac{3 x}{2} = 0$ or $\cos \frac{x}{2} = 0$

If $1 + \sin x = 0$ , then $\sin x = - 1$

$∴ x = 2 n π + \frac{3 π}{2}$

If $\cos \frac{3 x}{2} = 0$ , then $\frac{3 x}{2} = (2 n + 1) \frac{π}{2}$

$∴ x = (2 n + 1) \frac{π}{3}$

And if $\cos \frac{x}{2} = 0$ , then $\frac{x}{2} = (2 n + 1) \frac{π}{2}$

$∴ x = (2 n + 1) π$

But given interval is $[- π, π]$ .

Put $n = - 1$ in Eq. (i), $x = - \frac{π}{2}$

Put $n = 0, 1, - 1, - 2$ in Eq. (ii), $x = \frac{π}{3}, π - \frac{π}{3}, - π$

Hence, the solution in $[- π, π]$ are $- π, - \frac{π}{2}, - \frac{π}{3}, \frac{π}{3}, π$ .

Trigonometrical Equations 1 Question 30

30. Solve $2 (\cos x + \cos 2 x) + (1 + 2 \cos x) \sin 2 x$

Answer:

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Trigonometrical Equations 1 Question 30

30. Solve 2(cos⁡x+cos⁡2x)+(1+2cos⁡x)sin⁡2x

Answer:

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30. Solve $2 (\cos x + \cos 2 x) + (1 + 2 \cos x) \sin 2 x$