Question

The General solution of $9^{\sin x}-2.3^{\sin x}+1=0$ is

• A. $n \pi$
• B. $\frac{n \pi}{2}$
• C. $n \pi+(-1)^{n} \frac{\pi}{2}$
• D. $2 n \pi \pm \frac{\pi}{2}$

Answer 1

Answer: (A)

$9^{\sin x}-2.3^{\sin x}+1=0$
$\left(3^{\sin x}\right)^{2}-2.3^{\sin x}+1=0$
$\left(3^{\sin x}-1\right)^{2}=0$
$3^{\sin x}=1=3^{0}$
$\therefore \sin x=0$
$\therefore x=0^{\circ}, \propto=0^{\circ}$
The G.S is $x=n \pi+(-1)^{n} 0^{\circ}$
$\therefore x=n \pi$