## anonymous 3 years ago solve sin^2 x +2 sinx cosx = 0

1. anonymous

sinx(sinx+2cosx) = 0 What can you do now? think.

2. anonymous

if a*b = 0, Either a = 0 or b = 0 Here, a = sinx and b = sinx + 2cosx

3. anonymous

i don't think this is that easy

4. anonymous

that is $$\sin(x)=0$$ is easy enough to solve, you get $$x=n\pi, n\in \mathbb{Z}$$ but i can't see a good way to solve $\sin(x)=-2\cos(x)$

5. TheViper

$sin^2x+2\ sin\ x\ cos\ x=0$ $sin\ x(sin\ x+2\ cos\ x)=0$ $(sin\ x+2\ cos\ x)=0$

6. anonymous

that is not to say you cannot do it, but it is going to take some work

7. anonymous

the first thing you have to do to solve $\sin(x)+2\cos(x)=0$ is to write it as a single function of sine

8. anonymous

in general you can write $a\sin(x)+b\cos(x)$ as $\sqrt{a^2+b^2}\sin(x+\theta)$

9. anonymous

:'( i need help on mine

10. ParthKohli

I kinda wonder if you could do $$\sin^2(x) + \sin(2x)=0$$.

11. anonymous

in this case you get $\sin(x)+2\cos(x)=\sqrt{5}\sin(x+\theta)$ where $$\tan(\theta)=2$$

12. anonymous

tanx = -2 x = atan(-2) + n*(pi)

13. anonymous

therefore $$x+\theta=n\pi$$ gives $$\theta=n\pi-\tan^{-1}(2)$$

14. anonymous

you can check that it is right by looking here http://www.wolframalpha.com/input/?i=sin%28x%29%2B2cos%28x%29

15. anonymous

@ParthKohli i don't think that identity is going to help in this case you will have two arguments, one of $$x$$ and the other of $$2x$$ if you have a solution from that approach i would love to see it

16. TheViper

oh I had forgotten this site Thanx @satellite73 for remembering :)

17. ParthKohli

Gaahh @satellite73 you were right.