# Solving a trigonometric equation --> help needed!

• Jun 5th 2009, 01:33 PM
s3a
Solving a trigonometric equation --> help needed!
Can someone help me with fully illustrated steps for the following please?:

Solve the following trigonometric equation. Give the exact value(s).

2 sin^2 (x) - 3 cos (x) = 3 x ∈ [0, π]

Any help would be greatly appreciated!
• Jun 5th 2009, 01:56 PM
Amer
Quote:

Originally Posted by s3a
Can someone help me with fully illustrated steps for the following please?:

Solve the following trigonometric equation. Give the exact value(s).

2 sin^2 (x) - 3 cos (x) = 3 x ∈ [0, π]

Any help would be greatly appreciated!

$2(1-cos^2(x))-3cos(x)=3$

$2-2cos^2(x)-3cos(x)-3=0$

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

$(-2cosx-1)(cosx+1)=0$

so the first is zero or the second find when the first zero and when the second zero
• Jun 5th 2009, 01:59 PM
Soroban
Hello, s3a!

Quote:

Solve the following trigonometric equation. Give the exact value(s).

. . $2\sin^2\!x - 3\cos x \:=\:3 \qquad x \in [0, \pi]$

$\text{We have: }\;2\underbrace{\sin^2\!x} - 3\cos x \;=\;3$
. . . . . $2\overbrace{(1-\cos^2\!x)} - 3\cos x \;=\;3$

. . which simplifies to: . $2\cos^2\!x + 3\cos x + 1 \:=\:0$

. . which factors: . $(\cos x + 1)(2\cos x + 1) \:=\:0$

And we have two equations to solve:

. . $\cos x +1\:=\:0 \quad\Rightarrow\quad \cos x \:=\:-1 \quad\Rightarrow\quad\boxed{ x \:=\:\pi}$

. . $2\cos x + 1 \:=\:0 \quad\Rightarrow\quad \cos x \:=\:-\tfrac{1}{2} \quad\Rightarrow\quad\boxed{ x \:=\:\tfrac{2\pi}{3}}$