Question

$\text{Solve}\text{2co}{\text{s}}^{\text{2}}\text{x + 3sinx = 0}$

Answer 1

Solve: $2co{s}^{2}x+3sinx=0$

$2[1-si{n}^{2}x]+3sinx=0$

$2-2si{n}^{2}x+3sinx=0$

$2si{n}^{2}x-3sinx-2=0$

Let $sinx=y$

$2y^2-3y-2=0$

$2y^2-4y+y-2=0$

$2y(y-2)+1(y-2)=0$

$(2y+1)(y-2)=0$

$y=-\frac{1}{2};2$

$\therefore sinx=-\frac{1}{2};sinx=2$ not possible.

sin is negative in third and fourth quadrant

$x=\pi +\frac{\pi }{6};2\pi -\frac{\pi }{6}$

$x=\frac{7\pi }{6};\frac{11\pi }{6}$