Question

Equation $6{\sin }^2\Theta -5\sin \Theta +1=0$ is satisfied by

• A. $\Theta =\frac{\pi }{2}$
• B. $\Theta =\frac{\pi }{3}$
• C. $\Theta =\frac{\pi }{4}$
• D. $\Theta =\frac{\pi }{6}$

Answer 1

Answer: (D)

$\Theta =\frac{\pi }{6}$

Consider the given equation,

$6{\sin }^2\theta -5\sin \theta +1=0$

Let,$x=\sin \theta$

$6x^2-5x+1=0$

$6x^2-3x-2x+1=0$

$3x(2x-1)-1(2x-1)=0$

$(2x-1)(3x-1)=0$

Now,

$(2\sin \theta -1)(3\sin \theta -1)=0$

When,

$3\sin \theta -1=0$

$\sin \theta =\frac{1}{3}$

$\theta ={\sin }^{-1}\left(\frac{1}{3}\right)$

When,

$2\sin \theta -1=0$

$\sin \theta =\frac{1}{2}$

$\theta =\frac{\pi }{6}$

Hence, this is the answer.