Trig equations: cos4x=sin2x

**RAz** · March 19th 2010, 06:48 PM

I have to solve cos4x=sin2x where x is an element of 0, 360.

The problem, for me, is cos4x. I haven't proved this myself, but a quick search gave me that cos4x= $8cos^4x-8cos^2x+1$

By going along the lines of $8cos^4x-8cos^2x+1=2sinxcosx$ (see original question) I managed to arrive at sinx= $\frac{1}{3}$

This does not give an exact value and I was wondering if I had done this correctly. Thanks!

**Prove It** · March 19th 2010, 07:05 PM

Originally Posted by RAz

I have to solve cos4x=sin2x where x is an element of 0, 360.

The problem, for me, is cos4x. I haven't proved this myself, but a quick search gave me that cos4x= $8cos^4x-8cos^2x+1$

By going along the lines of $8cos^4x-8cos^2x+1=2sinxcosx$ (see original question) I managed to arrive at sinx= $\frac{1}{3}$

This does not give an exact value and I was wondering if I had done this correctly. Thanks!

$\cos{4x} = \sin{2x}$

$\cos^2{2x} - \sin^2{2x} = \sin{2x}$

$1 - \sin^2{2x} - \sin^2{2x} = \sin{2x}$

$1 - 2\sin^2{2x} = \sin{2x}$

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

You now have a quadratic equation in $\sin{2x}$ .

So let $\sin{2x} = X$ , so that you can see the Quadratic:

$0 = 2X^2 + X - 1$

$0 = 2X^2 + 2X - X - 1$

$0 = 2X(X + 1) - 1(X + 1)$

$0 = (X + 1)(2X - 1)$ .

So $X = -1$ or $X = \frac{1}{2}$ .

This means $\sin{2x} = -1$ or $\sin{2x} = \frac{1}{2}$ .

Case 1:

$\sin{2x} = -1$

$2x = 270^{\circ} + 360^{\circ}n$ , where $n$ is an integer.

$x = 135^{\circ} + 180^{\circ}n$ , where $n$ is an integer.

Case 2:

$\sin{2x} = \frac{1}{2}$

Since sine is positive in the first and second quadrants:

$2x = \left\{30^{\circ}, 180^{\circ} - 30^{\circ} \right\} + 360^{\circ} n$

$2x = \left\{ 30^{\circ}, 150^{\circ}\right\} + 360^{\circ} n$

$x = \left \{ 15^{\circ}, 75^{\circ} \right\} + 180^{\circ} n$ .

So putting it together:

$x = \left\{ 15^{\circ}, 75^{\circ}, 135^{\circ}\right\} + 180^{\circ}n$ .

Upon substituting $n = 0$ and $n = 1$ , we find all the possible solutions in the domain $0^{\circ} \leq x \leq 360^{\circ}$ :

$x = \left\{ 15^{\circ}, 75^{\circ}, 135^{\circ}, 195^{\circ}, 255^{\circ}, 315^{\circ} \right\}$ .

**RAz** · March 19th 2010, 10:13 PM

Thanks a lot, I appreciate it. Very clear.

**running-gag** · March 20th 2010, 01:08 AM

Just to give an alternative way

$\cos{4x} = \sin{2x}$

$\cos{4x} = \cos{(90^{\circ}-2x)}$

Case 1:
$4x = 90^{\circ}-2x+ 360^{\circ} n$

$6x = 90^{\circ} + 360^{\circ} n$

$x = 15^{\circ} + 60^{\circ} n$

$x = \left\{ 15^{\circ}, 75^{\circ}, 135^{\circ}, 195^{\circ}, 255^{\circ}, 315^{\circ} \right\}$

Case 2:
$4x = -(90^{\circ}-2x) + 360^{\circ} n$

$4x = -90^{\circ}+2x + 360^{\circ} n$

$2x = -90^{\circ} + 360^{\circ} n$

$x = -45^{\circ} + 180^{\circ} n$

$x = \left\{135^{\circ}, 315^{\circ} \right\}$

Therefore:
$x = \left\{ 15^{\circ}, 75^{\circ}, 135^{\circ}, 195^{\circ}, 255^{\circ}, 315^{\circ} \right\}$

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