Question

Solve the following equations.
$sin(2x+\frac{5\pi }{2})-3cos(\frac{7\pi }{2}-x)=1+2sinx.$

Answer 1

$sin(2x+\frac{5\pi }{2})-3cos(\frac{7\pi }{2}-x)=1+2sinx$

$sin(\theta +\frac{7\pi }{2})=cos\theta$

$cos(\frac{7\pi }{2}-\theta )=+sin\theta$

$\Rightarrow sin(2x+\frac{5\pi }{2})=cos2x$

$cos(\frac{7\pi }{2}-x)=sinx$

$\therefore cos2x-3sinx=1+2sinx$

$\Rightarrow 1-2si{n}^{2}x-3sinx=1+2sinx$

$\Rightarrow 2si{n}^{2}x+5sinx=0$

$\Rightarrow sinx=0$

$x=7\pi (n=0,\pm 1,\pm 2...)$