trig general solution

**srirahulan** · September 5th 2012, 07:17 AM

$cosx-sin3x=\sqrt2$
$6cos^2x+8sinxcosx=0$

How can I solve this equation and find out the general solution.Please give me any idea.

**Prove It** · September 5th 2012, 07:26 AM

Originally Posted by srirahulan

$cosx-sin3x=\sqrt2$
$6cos^2x+8sinxcosx=0$

How can I solve this equation and find out the general solution.Please give me any idea.

The second factorises very easily. Set each factor equal to 0.

**srirahulan** · September 5th 2012, 07:40 AM

Ok Thanks,how can I solve first one?

**kalyanram** · September 5th 2012, 08:21 AM

Write $cosx=sin(\frac{\pi}{2}-x)$ and use trigonometric transformation the equation becomes $cos(\frac{\pi}{4}+x)sin(\frac{\pi}{4}-2x)=\frac{1}{\sqrt{2}}$
Look for sign changes of this function in the interval $[0,2\pi]$ the roots are as shown in this link cos(pi/4+x)sin(pi/4 -2x) - 1/sqrt(2) - Wolfram|Alpha

**happyyan** · November 1st 2012, 12:37 AM

This is one of the best techniques to solving problems (not just mathematical by the way, but problems in general) and this is the first thing you should do any of these sorts of geometry problems. nike free running nike free running shoes for women nike free 3.0

**Salahuddin559** · November 1st 2012, 09:38 PM

Or, you will get a cubic in Sin^2(x) or cos^2(x), solve it using Cubic function - Wikipedia, the free encyclopedia.

Salahuddin
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