Trigonometric Functions — MATHS STD 11 Science — Question

$\Rightarrow2(1-\cos^{2}\text{x})=3\cos\text{x}$

$\Rightarrow2\cos^{2}\text{x}+3\cos\text{x}-2=0$

$\Rightarrow(2\cos\text{x}-1)(\cos\text{x}+2)=0$

$\Rightarrow\cos\text{x}=\frac{1}{2}$ or $\cos\text{x}=-2$

But, $\cos\text{x}=-2$ is not possible. $(-1\leq\cos\text{x}\leq1)$

 $\therefore\cos\text{x}=\frac{1}{2}=\cos\frac{\pi}{3}$

$\Rightarrow\text{x}=2\text{n}\pi\pm\frac{\pi}{3},\text{n}\in\text{z}$

Putting n=0 and n=1, We get

$\text{x}=\frac{\pi}{3},\frac{5\text{x}}{3}$ $(0\leq\text{x}\leq2\pi)$