# Thread: Solving a trig equation

1. ## Solving a trig equation

Hi,

The question is solve for ipi =< x <= pi:

$2sin(x)cos(x) + sin(x) = 0$

but i get as far as

$2cos^{3}(x) + cos^{2}(x) - 2cos(x) - 1 = 0$

and then can't get further. Am i on the right track?

2. Originally Posted by aceband
Hi,

The question is solve for ipi =< x <= pi:

$2sin(x)cos(x) + sin(x) = 0$

but i get as far as

$2cos^{3}(x) + cos^{2}(x) - 2cos(x) - 1 = 0$

and then can't get further. Am i on the right track?
It may work but not the way I'd go. I'd factorise the equation

$sin(x)(2cos(x)+1) = 0$

Either $sin(x)=0$ or $cos(x) = -\frac{1}{2}$

3. Originally Posted by aceband
Hi,

The question is solve for ipi =< x <= pi:

$2sin(x)cos(x) + sin(x) = 0$

but i get as far as

$2cos^{3}(x) + cos^{2}(x) - 2cos(x) - 1 = 0$

and then can't get further. Am i on the right track?
No.

You're meant to factorise:

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

Then use the null factor law to get two simple equations that are easily solved.