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Question: The general solution of \(0 < 4\theta < 2\pi\), for any integer n is....

The general solution of 0<4θ<2π0 < 4\theta < 2\pi, for any integer n is.

A

θ=7π24,11π24\theta = \frac{7\pi}{24},\frac{11\pi}{24}

B

(α+β)=0⇒cos⁡(α+β)=0(\alpha + \beta) = 0 \Rightarrow \cos(\alpha + \beta) = 0

C

α+β=(2n+1)π2,n∈I\alpha + \beta = (2n + 1)\frac{\pi}{2},n \in I

D

sin⁡(α+2β)=sin⁡(2α+2β−α)\sin(\alpha + 2\beta) = \sin(2\alpha + 2\beta - \alpha)

Answer

(α+β)=0⇒cos⁡(α+β)=0(\alpha + \beta) = 0 \Rightarrow \cos(\alpha + \beta) = 0

Explanation

Solution

=k{Σsin⁡(B+C)sin⁡(B−C)}= k\{\Sigma\sin(B + C)\sin(B - C)\}

⇒ =k{Σ12(cos⁡2C−cos⁡2B)}=0= k\left\{ \Sigma\frac{1}{2}(\cos 2C - \cos 2B) \right\} = 0 .