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Question: The solution of the equation \(\Rightarrow\) is....

The solution of the equation ⇒\Rightarrow is.

A

(1+cos⁡3x)+1−cos⁡(2x−2π3)=0(1 + \cos 3x) + 1 - \cos\left( 2x - \frac{2\pi}{3} \right) = 0

B

⇒\Rightarrow

C

2cos⁡23x2+2sin⁡2(x−π3)=02\cos^{2}\frac{3x}{2} + 2\sin^{2}\left( x - \frac{\pi}{3} \right) = 0

D

None of these

Answer

(1+cos⁡3x)+1−cos⁡(2x−2π3)=0(1 + \cos 3x) + 1 - \cos\left( 2x - \frac{2\pi}{3} \right) = 0

Explanation

Solution

2sin⁡2θ−3sin⁡θ−2=02\sin^{2}\theta - 3\sin\theta - 2 = 0

⇒ ⇒\Rightarrow

Squaring both sides, we get (2sin⁡θ+1)(sin⁡θ−2)=0(2\sin\theta + 1)(\sin\theta - 2) = 0

Where sin⁡θ=−12\sin\theta = - \frac{1}{2} or ∵sin⁡θ≠2)\because\sin\theta \neq 2).

∴\therefore sin⁡θ=sin⁡(−π6)\sin\theta = \sin\left( \frac{- \pi}{6} \right) or ⇒\Rightarrow as θ=nπ+(−1)n(−π6)⇒θ=nπ+(−1)n+1π6\theta = n\pi + ( - 1)^{n}\left( \frac{- \pi}{6} \right) \Rightarrow \theta = n\pi + ( - 1)^{n + 1}\frac{\pi}{6}

or ⇒\Rightarrow

θ=nπ+(−1)n7π6\theta = n\pi + ( - 1)^{n}\frac{7\pi}{6} {∵−π6is equivalent to 7π6}\left\{ \because\frac{- \pi}{6}\text{is equivalent to }\frac{7\pi}{6} \right\} or 3+1=rcos⁡α\sqrt{3} + 1 = r\cos\alpha.