Question

Find the general solution of the equation:
4 sin x cos x + 2 sin x + 2 cos x + 1 = 0

Answer 1

Given,

$4\text{sin x cos x}+2\text{sin x}+2\text{cos}+1=0$

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

$\Rightarrow (2sinx+1)(2cosx+1)=0$

$\text{Either}2sinx+1=0\text{or}2cosx+1=0$

$\Rightarrow sinx=\frac{-1}{2}\text{or}cosx=\frac{-1}{2}$

Now, if $\text{sin x}=(-\frac{1}{2})\Rightarrow \text{sin x}=\text{sin}(\pi +\frac{\pi }{6})=\text{sin}\frac{7\pi }{6}$

The general solution of this equation is:

$x=nx+(-1{)}^n\left(\frac{7\pi }{6}\right)$...(i)

and if $\text{cos x}=-\frac{1}{2}\Rightarrow \text{cos x}=\text{cos}(\pi -\frac{\pi }{3})=\text{cos}\frac{2\pi }{3}$

The general solution of this equation is

$x=2n\pi \pm \frac{2\pi }{3}$

$\Rightarrow x=2\pi (n\pm \frac{1}{3})$

$x=n\pi +(-1{)}^n\frac{7\pi }{6}\text{and x}=2n\pi \pm \frac{2\pi }{3}$