Factor 2sin^2 x + 3cos x – 3 = 0

[Liquidice_69]

im trying to solve this trig equation and the book says to factor it i think but its not clear, could anyone help. thxs

2sin^2 x + 3cos x – 3 = 0

 

[whozum]

$$sin^2x = 1 - cos^2x$$ is all you need. From there its a quadratic.

 

[Liquidice_69]

but how do you get the cos^2

 

[dextercioby]

Whozum gave you where that cos^2 comes from.U'd be using the fundamental identity of circular trigonometry

$$\sin^{2}x+\cos^{2}x\equiv 1 \ , \forall \ x\in\mathbb{R}$$

Daniel.

 

[Liquidice_69]

ok here's what i got so far

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

and i know how to factor
$$2x^2-3x+1=0$$
$$(2x-1)(x-1)=0$$
$$x=1, x=.5$$

but how do i solve with the cos

 

[dextercioby]

Well u get the equations

$$a) \ \cos x=1$$

$$b) \ \cos x=\frac{1}{2}$$

Fix a domain where u want to solve these equations.

Daniel.

 

[Liquidice_69]

ok, i want to solve it for negative pi to pi

 

[dextercioby]

Well,then i can tell u that the fist equation has only solution if the interval is $\left(-\pi,\pi\right)$,and the second has 2.

Daniel.

 

[Liquidice_69]

i don't get it, do u have to subtract pi from the second equasion

 

[dextercioby]

$$\cos x=\frac{1}{2}$$ has the solution

$$x=\pm \frac{\pi}{3}$$ in the interval u were looking for the solution.

Daniel.

 

[Liquidice_69]

ok i get it the answers are

for the cos x=1 0
for the cos x=.5 pi/3 and 5 pi/3