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Question: If \(\sin x + \sin y + \sin z = - 3\), where \(x = y = z = \frac{3\pi}{2},\) , then x =...

If sin⁡x+sin⁡y+sin⁡z=−3\sin x + \sin y + \sin z = - 3, where x=y=z=3π2,x = y = z = \frac{3\pi}{2}, , then x =

A

x,y,z∈[0,2π]x,y,z \in \lbrack 0,2\pi\rbrack

B

sin⁡2θ=cos⁡θ⇒cos⁡θ=cos⁡(π2−2θ)\sin 2\theta = \cos\theta \Rightarrow \cos\theta = \cos\left( \frac{\pi}{2} - 2\theta \right)

C

⇒\Rightarrow

D

None of these

Answer

⇒\Rightarrow

Explanation

Solution

a=3,b=1,c=4a = \sqrt{3},b = 1,c = 4

a2+b2=3+1=4<c2,3cos⁡x+4sin⁡x=6a^{2} + b^{2} = 3 + 1 = 4 < c^{2},3\cos x + 4\sin x = 6 35cos⁡x+45sin⁡x=65\frac{3}{5}\cos x + \frac{4}{5}\sin x = \frac{6}{5} cos⁡(x−θ)=65\cos(x - \theta) = \frac{6}{5}

Hence [where θ=cos⁡−1(3/5)]\lbrack\text{where }\theta = \cos^{- 1}(3/5)\rbrack.