# Solving a trig equation

• Dec 8th 2009, 12:05 PM
aceband
Solving a trig equation
Hi,

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

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

but i get as far as

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

and then can't get further. Am i on the right track?
• Dec 8th 2009, 12:09 PM
e^(i*pi)
Quote:

Originally Posted by aceband
Hi,

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

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

but i get as far as

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

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

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

Either $\displaystyle sin(x)=0$ or $\displaystyle cos(x) = -\frac{1}{2}$
• Dec 8th 2009, 12:09 PM
mr fantastic
Quote:

Originally Posted by aceband
Hi,

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

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

but i get as far as

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

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

No.

You're meant to factorise:

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

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