Question

The general solution of the equation ${\sin }^2\theta ={\sin }^2\alpha$ is/are

• A.
  $n\pi +\alpha ,n\in Z.$
• B.
  $n\pi -\alpha ,n\in Z.$
• C.
  $(2n+1)\frac{\pi }{2}-\alpha ,n\in Z.$
• D.
  $(2n+1)\frac{\pi }{2}+\alpha ,n\in Z.$

Answer 1

Answer: (A), (B)

$n\pi -\alpha ,n\in Z.$
Here, in order to find the solution of the equation, we will express the equation in linear powers i.e.
we can write the equation as:
${\sin }^2\theta ={\sin }^2\alpha (\sin \theta -\sin \alpha )(\sin \theta +\sin \alpha )=04\times \left(\sin \frac{(\theta -\alpha )}{2}\cos \frac{(\theta +\alpha )}{2}\right)\left(\sin \frac{(\theta +\alpha )}{2}\cos \frac{(\theta -\alpha )}{2}\right)=0\sin (\theta -\alpha )\times \sin (\theta +\alpha )=0\sin (\theta -\alpha )=0or\sin (\theta +\alpha )=0(\theta -\alpha )=n\pi or(\theta +\alpha )=n\pi \Rightarrow \theta =n\pi +\alpha or\theta =n\pi -\alpha ,n\in Z\Rightarrow \theta =n\pi \pm \alpha ,n\in Z$