Math Help Forum - View Single Post - Solving a trigonometric equation --> help needed!

Soroban · #3 June 5th, 2009, 02:59 PM

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:

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

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