Trigonometric Equation

Given the function $2 \sin(2x+\frac{\pi}{2})$ find the amplitude, the period, the phase shift and the intersection points with the parent function, $\sin(x)$.

I was able to find the characteristics:

Period: $\pi$

Amplitude: 2

Phase shift: $\frac{\pi}{4}$

However I couldn't solve the equation $2\sin(2x+\frac{\pi}{2})=\sin(x)$. My difficulty was those factors that multiply the sine and the variabel.

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Using $\sin(a+b)=\sin a\cos b+\sin b\cos a$ gives you $$2\cos 2x=\sin x.$$ Now, using $\cos 2a=1-2\sin^2 a$ gives $$2(1-2\sin^2 x) =\sin x.$$ The above can be written as $$4\sin^2 x+\sin x-2=0.$$ This is a quadratic equation in $\sin x$. By the quadratic formula: $$\sin x={-1\pm\sqrt{33}\over 8 }.$$ Then, for $x$ in the interval $[-\pi/2,\pi/2]$ $$x=\arcsin({-1\pm\sqrt{33}\over 8 })$$ ($x\approx .63487$ or $x\approx -1.003$).
Both of these are solutions to the original equation $2\sin(2x+{\pi\over2})=\sin x$.
Thanks.I missed the sum identitie.I suppose than $a=2x$ and $b=\frac{\pi}{2}$. –  Pedro Nov 28 '11 at 10:56