Question

solve: sin^2 x + sin x cos x = 0

Answer 1

\[\sin ^{2} x + \sin x \cos x = 0\]

Answer 2

Factoring \(\sin(x)\) out, we get:
\[
\sin^2x+\sin x\cos x = (\sin(x))(\sin(x)+\cos(x))=0
\]Therefore, we have two solvable equations:
\[
\sin(x)=0
\]And
\[
\sin(x)=-\cos(x)
\]This turns into:
\[
\sin(x)=\sin(x-\frac{\pi}{2})
\]Solving each of these gives you the desired result.

Answer 3

Factor out sin(x):
\[\sin(x)(\sin(x)+\cos(x))=0 \implies \sin(x)=0 , \sin(x)+\cos(x)=0\]\[\sin(x)=0 \implies x = \pi k| k \in \mathbb{Z}\]\[\sin(x)+\cos(x)=0 \implies [\sin(x)+\cos(x)]^2=0 \]\[\implies \sin^2(x)+2\sin(x)\cos(x)+\cos^2(x)=0 \implies \sin(2x)+1=0\]\[\implies \sin(2x)=-1 \implies 2x =\frac{ 3 \pi }{ 2 } + 2\pi k \implies x =\frac{ 3 \pi }{ 4}+ \pi k|k \in \mathbb{Z}\] Do you understand?

@jdc1980

Answer 4

How did you get the sin (x-pi/2) ?

Answer 5

Note that \(\sin(x)\) is an odd function and that \(\cos(x)=\sin(\frac{\pi}{2}-x)\) by the sum of angles of a right triangle.
So, negating the above, we get:
\[
-cos(x)=-\sin(\frac{\pi}{2}-x)=\sin(-(\frac{\pi}{2}-x))=\sin(x-\frac{\pi}{2})
\]

Answer 6

Still not clicking... Is \[\csc x = \sin (\frac{ x }{ 2 } - x)\] a common identity?

Answer 7

cos sorry

Answer 8

It's a common identity, because the angles must add up to 90 degrees or \(\frac{\pi}{2}\), right? (Remember we're dealing with a right triangle). Hence \(\cos(x)=\sin(\frac{\pi}{2}-x\))

Answer 9

okay! I see! now I'm looking for all solutions where x is 0 <= x < pi

what should I do next?

Answer 10

So, we have that:
\[
\sin(\frac{3\pi}{4})=\sin(\frac{3\pi}{4}-\frac{\pi}{2})
\]Which is our only solution in the range \(x\in[0, \pi)\).

Answer 11

This is where I'm at: