Q.Solve the following equations for which solutions lies in the interval $0^\circ \le \theta < 360^\circ$:

Part (i), Step 1. Factor. $\sin^4x=\sin^2x \Rightarrow \sin^4x-\sin^2x=0 \Rightarrow \sin^2x(\sin^2x-1)=0$.

Part (i), Step 2. Solve each factor. Either $\sin^2x=0$ or $\sin^2x=1$.

$\sin^2x=0 \Rightarrow \sin x=0 \Rightarrow x=0^\circ,180^\circ$ (in $[0^\circ,360^\circ)$).

$\sin^2x=1 \Rightarrow \sin x=\pm1 \Rightarrow x=90^\circ$ (from $\sin x=1$) or $x=270^\circ$ (from $\sin x=-1$).

Part (i), Step 3. Collect. $x=0^\circ,90^\circ,180^\circ,270^\circ$. Check $x=90^\circ$: $\sin^4(90^\circ)=1=\sin^2(90^\circ)$ ✓.

Part (ii), Step 1. Rearrange into a quadratic in $\cos x$. $2\cos^2x+1=-3\cos x \Rightarrow 2\cos^2x+3\cos x+1=0$.

Part (ii), Step 2. Factor. $2\cos^2x+3\cos x+1=(2\cos x+1)(\cos x+1)=0$, so $\cos x=-\dfrac12$ or $\cos x=-1$.

Part (ii), Step 3. Solve each in $[0^\circ,360^\circ)$. $\cos x=-\dfrac12=\cos120^\circ \Rightarrow x=120^\circ,240^\circ$ (quadrants II and III). $\cos x=-1 \Rightarrow x=180^\circ$.

Part (ii), Step 4. Collect. $x=120^\circ,180^\circ,240^\circ$. Check $x=120^\circ$: $2\cos^2120^\circ+1=2(0.25)+1=1.5$; $-3\cos120^\circ=-3(-0.5)=1.5$ ✓.

Part (iii), Step 1. Rearrange into a quadratic in $\sin x$. $2\sin^2x+1=3\sin x \Rightarrow 2\sin^2x-3\sin x+1=0$.

Part (iii), Step 2. Factor. $2\sin^2x-3\sin x+1=(2\sin x-1)(\sin x-1)=0$, so $\sin x=\dfrac12$ or $\sin x=1$.

Part (iii), Step 3. Solve each in $[0^\circ,360^\circ)$. $\sin x=\dfrac12=\sin30^\circ \Rightarrow x=30^\circ,150^\circ$. $\sin x=1 \Rightarrow x=90^\circ$. …

Reduce to a quadratic in one ratio, factor, reject roots outside $[-1,1]$, …

• Forgetting that $\sin^2x=1$ gives TWO angles ($90^\circ$ and $270^\circ$), not just one …