How do you solve for 2sinxcosx=2sinx on the interval [0,2pi]

Question

anonymous · Answer

First divide both sides by $2\sin x$, and you get: $$
\cos x =1
$$This is your first equation.  Now, we can only do this when $$
2\sin x\neq 0
$$Now we we plug $2\sin x=0$ into the equation we get: $$
0 \cos x = 0\implies 0=0
$$ Which is a true statement so we need to solve $$
2\sin x = 0
$$as well

danjs · Answer

Cos(theta) = 1

 The cosine of an angle, is the X coordinate of the point (cos(t), sin(t)) on the unit circle. At what angles , is the X coordinate equal to 1 on the unit circle? Since the radius is 1, All of the radius is in the X direction, along the x axis. Zero Degrees

danjs · Answer

Same thought process for the second equation @wio mentioned above

anonymous · Answer

The 'proper way' is to subtract then factor: $$
2\sin x\cos x = 2\sin x\
2\sin x \cos x - 2\sin x =0\
2\sin x(\cos x-1)=0
$$It leads to the same equations.

anonymous · Answer

Thank you @wio and @DanJS