# Solving Trigonometric Equation

• Feb 20th 2013, 09:17 AM
vaironxxrd
Solving Trigonometric Equation
Hello, again, Everyone.

I have the following problem

$2sin^2x = 2 + cos x$

My attempt:

$2sin^2x = 2 + cos x$

$2sin^2x - cos x - 2 = 0$

$x = \frac{3pi}{6}$
• Feb 20th 2013, 09:43 AM
veileen
Re: Solving Trigonometric Equation
Hello! ^^

$2\sin^2x-\cos x-2=0\Rightarrow 2(1-\cos^2x)-\cos x-2=0$

$\Rightarrow 2-2\cos^2x-\cos x-2=0\Rightarrow 2\cos^2x+\cos x=0$

$\Rightarrow \cos x(2\cos x+1)=0 \Rightarrow$

1. $\cos x=0\Rightarrow x=\pm \frac{\pi}{2}+2k\pi$, $k \in \mathbb{Z}$

2. $\cos x=-\frac{1}{2}\Rightarrow x = \pm \frac{2\pi}{3}+2k\pi$, $k \in \mathbb{Z}$

$\Rightarrow x \in \left \{ \pm \frac{\pi}{2}+2k\pi \, |\, k \in \mathbb{Z} \right \}\cup \left \{ \pm \frac{2\pi}{3}+2k\pi \, |\, k \in \mathbb{Z} \right \}$

Those solutions are for $x \in \mathbb{R}$. If $x \in [0, 2\pi)$, then the solutions are $\frac{\pi}{2}, \frac{3\pi}{2}, \frac{2\pi}{3}, \frac{4\pi}{3}$.
• Feb 20th 2013, 09:47 AM
MINOANMAN
Re: Solving Trigonometric Equation
Vaironxxrd

transfer the 2 to the left side and then substitute -2 with the basic trigonometric identity (cosx)^2+(sinx)^2=1 to find 2(cosx)^2+cosx=0
then factorise to get cosx[2cosx+1]=0..then solve it it is easy..try....it
MINOAS