sin^2x=.5sinx
sinx=.5
x=pi/6
interval: 0

Question

sin^2x=.5sinx sinx=.5 x=pi/6 interval: 0

anonymous · Answer

it's quadratic and should have 2 roots.
tosolve you must subtract ,5sinx from both sides, factor and find the zeros.

anonymous · Answer

there are 2 roots in the interval

mathmale · Answer

sin^2x=.5sinx might be easier to understand and solve if written as 2(sin x)^2 = sin x.
Solve for sin x first.  And yes, there are TWO primary roots on the given interval.

anonymous · Answer

$$\sin ^2x-.5\sin x=0,\sin x \left( \sin x-.5 \right)=0$$

$$Either \sin x=0,x=?$$
$$or \sin x=0.5,x=?$$

anonymous · Answer

so is the error that it needs two primary roots on the interval?

mathmale · Answer

@15HReeves :  In your shoes, I'd solve this problem myself.  Then, I'd go back and compare my answers to the ones given.  Then I'd likely quickly spot any discrepancies.

mathmale · Answer

Worry about the error later.  Find the correct solutions now, and then compare your solutions with that / those  given.

anonymous · Answer

$$x^2=.5x$$ if you divide by $x$ and write 
$$x=.5$$ then you miss the obvious solution $x=0$