# trig equation

G

#### Guest

##### Guest

1. 2sin[squared]x+3cosx=0

#### happy

##### Full Member
Well, what are your thoughts on this one since a similar one has been done for you? Where are you stuck?

G

#### Guest

##### Guest

I guess the different signs threw me off.

#### pka

##### Elite Member
The problem can be rewritten as $$\displaystyle \begin{eqnarray*} 2\sin ^2 (x) + 3\cos (x) & = & 0 \\ 2 - 2\cos ^2 (x) + 3\cos (x) & = & 0 \\ 2\cos ^2 (x) - 3\cos (x) - 2 & = & 0 \\ \end{array}$$.

Now can you solve $$\displaystyle w = \cos (x)\quad \Rightarrow \quad 2w^2 - 3w - 2 = 0$$

G

(2w+1)(w-2)=0
2w+1=0 w-2=0

w=-1/2 w=2

#### tkhunny

##### Moderator
Staff member
OK, now we have:

cos(x) = -1/2

1) Why did I throw out W = 2?
2) What are the solutions for x?

G

#### Guest

##### Guest
tkhunny said:
OK, now we have:

cos(x) = -1/2

1) Why did I throw out W = 2?
2) What are the solutions for x?

because cos doesn't go to 2

Staff member
Excellent.