Solving for x

**db5vry** · March 10th 2010, 11:11 PM

This is part of a bigger question to do with polar coordinates - I have obtained a derivative and equated it to zero in order to indicate a stationary point and this is what I need to solve:

$2cos(2x)cos(x) - sin(2x)sin(x) = 0$

Which of the double/half/compound angle formulas do I need to use here? I always seem to forget in questions like these.

Thanks if you can help me :]

**Sudharaka** · March 10th 2010, 11:22 PM

Originally Posted by db5vry

This is part of a bigger question to do with polar coordinates - I have obtained a derivative and equated it to zero in order to indicate a stationary point and this is what I need to solve:

$2cos(2x)cos(x) - sin(2x)sin(x) = 0$

Which of the double/half/compound angle formulas do I need to use here? I always seem to forget in questions like these.

Thanks if you can help me :]

Dear db5vry,

Try, $sin2x=2sinxcosx~and~cos2x=2cos^{2}x-1=1-2sin^{2}x=cos^{2}x-sin^{2}x$

Hoe this will help you.

**db5vry** · March 10th 2010, 11:33 PM

Originally Posted by Sudharaka

Dear db5vry,

Try, $sin2x=2sinxcosx~and~cos2x=2cos^{2}x-1=1-2sin^{2}x=cos^{2}x-sin^{2}x$

Hoe this will help you.

Will this then lead me to

$cosx (6sin^2x - 2) = 0$ which gives the solutions $\frac{\pi}{2}$ and $\frac{\sqrt{3}}{3}$ ? If so then I've done this correctly, but I'm not sure!

**Sudharaka** · March 11th 2010, 12:21 AM

Originally Posted by db5vry

Will this then lead me to

$cosx (6sin^2x - 2) = 0$ which gives the solutions $\frac{\pi}{2}$ and $\frac{\sqrt{3}}{3}$ ? If so then I've done this correctly, but I'm not sure!

Dear db5vry,

You could eliminate your doubts by substituting the solutions in the original equation.

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