Sine/Cosine Equations

**Adam We** · February 21st 2009, 12:50 AM

I don't understand how to do these following equations:

Solve for x:

2(cos x)^2 = cos x for -(pi) < x < (pi)

cos 2x = 5sinx - 2 for all real x

I'm more interested in how to do them rather than the answers themselves.

Thank you very much, Adam

**running-gag** · February 21st 2009, 01:15 AM

Hi

$2 \cos^2 x = \cos x$

$\cos x\: (2 \cos x -1) = 0$

$\cos x = 0$ or $\cos x = \frac{1}{2}$

$\cos 2x = 5 \sin x - 2$

$1 - 2 \sin^2 x = 5 \sin x - 2$

$2 \sin^2 x + 5 \sin x - 3 = 0$

$2 X^2 + 5 X - 3 = 0$ with $X = \sin x$

**Air** · February 21st 2009, 01:17 AM

Originally Posted by Adam We

I don't understand how to do these following equations:

Solve for x:

2(cos x)^2 = cos x for -(pi) < x < (pi)

cos 2x = 5sinx - 2 for all real x

I'm more interested in how to do them rather than the answers themselves.

Thank you very much, Adam

$2(\cos x)^2 = \cos x \ \ \ \ -\pi<x<\pi$

Let $u=\cos x$ and rearrange to get it into a quadratic equation form.

$2u^2 - u=0$

This is in quadratic equation form as we have polynomial to power of

$2$ thus we can factorise and solve for solutions.
$u(2u-1) = 0$
$u = 0$ or $2u - 1= 0$

So your solutions will occur when $\cos x = 0$ and $\cos x = \frac{1}{2}$ .

As $\cos x$ is a periodic function, thus there may be more than one solution of $x$ for between $-\pi<x<\pi$ . One way to find the solution is to plot the graph between $-\pi<x<\pi$ for $\cos x$ and draw a line at $\frac{1}{2}$ and $0$ and see where it intercepts the curve. Those would be you solutions.

__________________
$\cos 2x = 5\sin x - 2$

You need to know the double angle rule. For $\cos 2x$ , there are three versions and you must pick the appropriate one. For your example, we would pick one involving $\sin x$ . This would be $\cos 2x = 1 - 2\sin ^2 x$ .

$\therefore 1 - 2\sin ^2 x = 5\sin x - 2$

Now you can let $u=\sin x$ and solve for solution of the quadratic equation and replace it with your substitution to find the specific values (as done in the first question).

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