The general solution of the equation $\sqrt{3-5 \sin x+\sin ^2 x}+\cos x=0$ is
The general solution of the equation $\sqrt{3-5 \sin x+\sin ^2 x}+\cos x=0$ is
- $n \pi+(-1)^n \frac{\pi}{6}, n \in Z$
- $2 n \pi \pm \frac{\pi}{6}, n \in Z$
- $(2 n+1) \pi-\frac{\pi}{6}, n \in Z$
- $2 n \pi \pm \frac{5 \pi}{6}, n \in Z$
Solution
Given equation,
$
\begin{gathered}
\sqrt{3-5 \sin x+\sin ^2 x}+\cos x=0 \\
\Rightarrow \quad \sqrt{3-5 \sin x+\sin ^2 x}=-\cos x,\{\cos x\}
\end{gathered}
$
On squaring both sides, we get
$
\begin{array}{lrl}
\Rightarrow & 3-5 \sin x+\sin ^2 x=\cos ^2 x \\
\Rightarrow & 2 \sin ^2 x-5 \sin x+2=0 \\
\Rightarrow & 2 \sin ^2 x-4 \sin x-\sin x+2=0 \\
\Rightarrow & 2 \sin x(\sin x-2)-1(\sin x-2)=0 \\
\Rightarrow & \sin x=\frac{1}{2} \text { or } 2 \\
\because & \sin x \in[-1,1] \\
\therefore & \sin x=\frac{1}{2} \text { and } \cos x < 0
\end{array}
$
Means, $\quad \sin x=\frac{1}{2}$ and $\cos x=\frac{-\sqrt{3}}{2}$ Therefore, $x=(2 n+1) \pi-\frac{\pi}{6}, n \in Z$
Asked in: AP EAMCET 2018 (22 Apr Shift 2)
Practice more Trigonometric Equations questions on Aicharya