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Explanation:

Step1: Simplify the first - equation

Use the trigonometric identities. We know that $\sin(A + B)=\sin A\cos B+\cos A\sin B$ and $\cos(A + B)=\cos A\cos B-\sin A\sin B$. For $\sin(x+\frac{3\pi}{2})=-\cos x$ and $\cos(x + \frac{\pi}{2})=-\sin x$. The equation $\sin(x+\frac{3\pi}{2})-\cos(x+\frac{\pi}{2}) = 1$ becomes $-\cos x+\sin x = 1$, or $\sin x-\cos x = 1$. Square both sides: $(\sin x-\cos x)^2=1^2$. Expanding gives $\sin^{2}x - 2\sin x\cos x+\cos^{2}x = 1$. Since $\sin^{2}x+\cos^{2}x = 1$, the equation simplifies to $1 - 2\sin x\cos x=1$, so $\sin x\cos x = 0$. If $\sin x = 0$, then $x = 0,\pi,2\pi$ (but we are in the interval $[0,2\pi)$ so we consider $x = 0,\pi$). Substituting into $\sin x-\cos x = 1$, when $x = 0$, $\sin(0)-\cos(0)=- 1\neq1$, when $x=\pi$, $\sin(\pi)-\cos(\pi)=1$. If $\cos x = 0$, then $x=\frac{\pi}{2},\frac{3\pi}{2}$. When $x=\frac{\pi}{2}$, $\sin(\frac{\pi}{2})-\cos(\frac{\pi}{2})=1$.

Step2: Simplify the second - equation

The equation $7\cos(x-\frac{17\pi}{2})=7\sin^{2}x$. We know that $\cos(A - B)=\cos A\cos B+\sin A\sin B$. Also, $\cos(x-\frac{17\pi}{2})=\cos(x - 8\pi-\frac{\pi}{2})=\cos(x-\frac{\pi}{2})=\sin x$. The equation becomes $\sin x=\sin^{2}x$, or $\sin^{2}x-\sin x = 0$. Factor out $\sin x$: $\sin x(\sin x - 1)=0$. If $\sin x = 0$, then $x = 0,\pi$ in the interval $[0,2\pi)$. If $\sin x = 1$, then $x=\frac{\pi}{2}$.

Answer:

For the first equation: $\frac{\pi}{2},\pi$; For the second equation: $0,\frac{\pi}{2},\pi$