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Please help me understand how to solve this for $0\leq x\leq360 $

I seem to have a problem with equations with powers.

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

thinking that I would start by simplifying:

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

How I wish the equation in the bracket was in form $\sin^2 x+ \cos^2x$ which is equal to 1.

I also tried to substitute $\sin^2 x=1- \cos^2x$

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Is $\cos^{x}$ or $\cos x$ ?? –  Wmmoreno Nov 1 '13 at 3:48
    
@Wmmoreno Please don't remove all of the OP's work when editing a question. –  T. Bongers Nov 1 '13 at 3:51
    
@T.Bongers ok Sorry ... XD –  Wmmoreno Nov 1 '13 at 3:52

3 Answers 3

up vote 2 down vote accepted

Well since you know $sin^2{x} = 1 - cos^2{x}$ then let's try putting it into the equation: \begin{eqnarray} 3(1 - 2cos^2{x}) + cosx - 1& & \end{eqnarray} Now try to substitute $u = cos(x)$, we get then $3(1 - 2u^2) + u - 1 = -6u^2 + u + 2$ now solve for $u$. Don't forget that $u = cos(x)$, so $x = arccos(u)$ .

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You shouldn't reuse $x$ in $x=\cos (x)$, but otherwise this is a good approach. It responds well to OP's claim of trying $\sin^2 x = 1-\cos^2 x$ –  Ross Millikan Nov 1 '13 at 3:52
    
Oh you make a really good point :P ... I will edit right now. –  InsigMath Nov 1 '13 at 3:53
    
looks great! I am studying it, to understand the logic –  Sylvester Nov 1 '13 at 4:02
    
Multiple layers of substitutions is a common technique to get comfortable with, don't worry about not seeing it right away. Takes practice :) –  InsigMath Nov 1 '13 at 4:09
    
@InsigMath and Ross Millikan: Here comes my question: The original equation asks us to solve the equation. What are we solving? for x? If the final solution is $cosx = {2/3} $ Doesn't this come out like I was solving for $cosx$ –  Sylvester Nov 1 '13 at 4:27

$3\sin^2{x} - 3\cos^2{x} + \cos{x} - 1 = 0$

$6\cos^2{x} - \cos{x} - 2 = 0$

$\cos{x} = -\frac{1}{2}\:\:\:\cos{x} = \frac{2}{3}$

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but how to find $\arccos \frac23$ in $[0,360^\circ]?$ –  lab bhattacharjee Nov 1 '13 at 3:59
    
It will just be $\arccos{\frac{2}{3}}$ and $2\pi - \arccos{\frac{2}{3}}$ –  LTS Nov 1 '13 at 4:04
    
I think what @labbhattacharjee may be asking, is why the $2/3$ and not $-1/2$. Well it is because we are looking for positive values only in range 0-360. –  Sylvester Nov 1 '13 at 4:30
    
@Sylvester, no no. My point is that $\frac23$ is not known standard value for $\cos x$ –  lab bhattacharjee Nov 1 '13 at 4:32
    
@labbhattacharjee, I understand. Thanks. –  Sylvester Nov 1 '13 at 4:37

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

$6\cos^2 x-\cos x - 2 = 0$

By the quadratic formula

$cos x = \dfrac{1 \pm \sqrt{1-4(6)(-2)}}{12} =\dfrac{-1\pm 7}{12}$

So $cos x = \frac{-1}{2}$ and $cos x = \frac{2}{3}$

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