Question

If $n$ be the number of solution of the equation $3cos^{2}x-8sinx=0$ in $\left[0 , 3 \pi \right]$ is, then the value of $n$ is equal to

Answer 1

We have,
$ \begin{array}{l} 3 \cos ^{2} x-8 \sin x=0 \\ \Rightarrow 3\left(1-\sin ^{2} x\right)-8 \sin x=0 \\ \Rightarrow 3-3 \sin ^{2} x-8 \sin x=0 \\ \Rightarrow 3 \sin ^{2} x+8 \sin x-3=0 \\ \Rightarrow 3 \sin ^{2} x+9 \sin x-\sin x-3=0 \\ \Rightarrow 3 \sin x(\sin x+3)-(\sin x+3)=0 \\ \Rightarrow(3 \sin x-1)(\sin x+3)=0 \\ \Rightarrow \sin x=\frac{1}{3}, -3 \end{array} $
But $\sin x \in[-1,1]$ for all $x \in R$. Hence,
$ \sin x=\frac{1}{3} $