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The question is:

2 cos^2(x)-sin(x)-1 = 0

I need to solve the equation for all values of x between 0 < x < 360

Now I got it to here and got a bit stuck:

2 cos^2(x)+cos(x)-2 = 0

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Linear algebra is concerned with vector spaces of all dimensions and linear transformations between them. – Don Larynx Oct 9 '13 at 0:09
HINT: Quadratic equation. – Don Larynx Oct 9 '13 at 0:10
@DonLarynx Thanks! Yeah, I got to the the quadratic equation above however it wasnt really factoring down. – Joseph Wright Oct 9 '13 at 0:11
First equation isn't equivalent to second one. You should replace $\cos^2(x)$ into $1-\sin^2(x)$. 2nd equation will become as $2(1-\sin^2(x))-\sin(x)-1=0$, or $2\sin^2(x)+\sin(x)-1=0$. – Oleg567 Oct 9 '13 at 0:34
up vote 0 down vote accepted

Quadratic equation: $$cos(x) = \frac{-1 \pm \sqrt{17}}{2}$$

Can you continue?

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I got to there. So all I need to do is arccos each of the x values and then get their values in degrees? Because thats what I thought I needed to do but im second guessing myself now:L – Joseph Wright Oct 9 '13 at 0:13
Yes................ – Don Larynx Oct 9 '13 at 0:23

Starting equation: $$ 2\cos^2(x)-\sin(x)-1=0 \tag{1} $$

Applying identity $$ \sin^2(x)+\cos^2(x)=1, $$ you'll transform equation $(1)$ to form $$ 2(1-\sin^2(x)) - \sin(x)-1=0; $$ $$ -2\sin^2(x) - \sin(x)+1=0; $$ $$ 2\sin^2(x) + \sin(x)-1=0. \tag{2} $$

Last equation is quadratic equation on $\sin(x)$.

There are $2$ solutions of quadratic equation $(2)$:

$$ \sin(x)=\dfrac{1}{2}, ~~~~~~~~~~\sin(x)=-1. $$

So, there are $3$ solutions of trigonometric equation $(1)$, when $0<x<360^\circ$. Find them all...

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