## shaqadry Group Title solve sin^2 x +2 sinx cosx = 0 one year ago one year ago

1. beknazar23 Group Title

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

2. beknazar23 Group Title

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

3. satellite73 Group Title

i don't think this is that easy

4. satellite73 Group Title

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 Group Title

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

6. satellite73 Group Title

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

7. satellite73 Group Title

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. satellite73 Group Title

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

9. tayluvsya Group Title

:'( i need help on mine

10. ParthKohli Group Title

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

11. satellite73 Group Title

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

12. beknazar23 Group Title

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

13. satellite73 Group Title

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

14. satellite73 Group Title

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

15. satellite73 Group Title

@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 Group Title

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

17. ParthKohli Group Title

Gaahh @satellite73 you were right.