Question

solve sin^2 x +2 sinx cosx = 0

Answer 1

sinx(sinx+2cosx) = 0

What can you do now? think.

Answer 2

if a*b = 0,
Either a = 0 or b = 0

Here, a = sinx and b = sinx + 2cosx

Answer 3

i don't think this is that easy

Answer 4

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
\[\sin(x)=-2\cos(x)\]

Answer 5

\[sin^2x+2\ sin\ x\ cos\ x=0\]
\[sin\ x(sin\ x+2\ cos\ x)=0\]
\[(sin\ x+2\ cos\ x)=0\]

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

:'( i need help on mine

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

@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

Answer 16

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

Answer 17

Gaahh @satellite73 you were right.