$\displaystyle cosx-sin3x=\sqrt2$
$\displaystyle 6cos^2x+8sinxcosx=0$
How can I solve this equation and find out the general solution.Please give me any idea.
Write $\displaystyle cosx=sin(\frac{\pi}{2}-x)$ and use trigonometric transformation the equation becomes $\displaystyle cos(\frac{\pi}{4}+x)sin(\frac{\pi}{4}-2x)=\frac{1}{\sqrt{2}}$
Look for sign changes of this function in the interval $\displaystyle [0,2\pi]$ the roots are as shown in this link cos(pi/4+x)sin(pi/4 -2x) - 1/sqrt(2) - Wolfram|Alpha
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
Or, you will get a cubic in Sin^2(x) or cos^2(x), solve it using Cubic function - Wikipedia, the free encyclopedia.
Salahuddin
Maths online