Question

The value of $\alpha$ satisfying $6{\sin }^2\alpha -5\cos \alpha =2$ is

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

Answer 1

Answer: (B)

$\alpha =\frac{\pi }{3}$

Consider the given equation,

$6{\sin }^2\alpha -5\cos \alpha =2$

$6(1-{\cos }^2\alpha )-5\cos \alpha =2$

$6-6{\cos }^2\alpha -5\cos \alpha -2=0$

$-6{\cos }^2\alpha -5\cos \alpha +4=0$

$6{\cos }^2\alpha +5\cos \alpha -4=0$

$6{\cos }^2\alpha +8\cos \alpha -3\cos \alpha -4=0$

$2\cos \alpha (3\cos \alpha +2)-1(3\cos \alpha +2)=0$

$(3\cos \alpha +2)(2\cos \alpha -1)=0$

When,

$3\cos \alpha +2=0$

$\cos \alpha =\frac{-2}{3}$

$\alpha ={\cos }^{-1}\left(\frac{-2}{3}\right)$

When,

$2\cos \alpha -1=0$

$\cos \alpha =\frac{1}{2}$

$\cos \alpha =\cos \frac{\pi }{3}$

$\alpha =\frac{\pi }{3}$

Hence, this is the answer.