Trigonometry (11th Edition) Clone

The solution set is $$\theta=\{22.5^\circ+n90^\circ,n\in Z\}$$
$$4\cos2\theta=8\sin\theta\cos\theta$$ - Recall the identity: $2\sin\theta\cos\theta=\sin2\theta$ Therefore, $$8\sin\theta\cos\theta=4\sin2\theta$$ Apply back to the equation: $$4\cos2\theta=4\sin2\theta$$ $$\cos2\theta=\sin2\theta$$ Here we can divide both sides by $\cos2\theta$ to get $\tan2\theta$. However, we need to prove first that $\cos2\theta\ne0$ in this situation. 1) Prove that $\cos2\theta\ne0$ If $\cos2\theta=0$, then $\sin2\theta=\cos2\theta=0$ However, we know that there are no such values of $\theta$ that have both $\sin2\theta=\cos2\theta=0$ Therefore, $\cos2\theta\ne0$ 2) Divide both sides by $\cos2\theta$ $$1=\frac{\sin2\theta}{\cos2\theta}$$ According to the identity $\tan\theta=\frac{\sin\theta}{\cos\theta}$, we have $$\tan2\theta=1$$ 3) Solve the equation over the interval $[0^\circ,360^\circ)$ Over the interval $[0,2\theta)$, there are 2 values of $2\theta$ where $\tan2\theta=1$, which are $\{45^\circ,225^\circ\}$ Therefore, $$2\theta=\{45^\circ,225^\circ\}$$ 2) Solve the equation for all solutions Tangent function has period $180^\circ$, so we would add $180^\circ$ to all solutions found in part 1) for $2\theta$. $$2\theta=\{45^\circ+n180^\circ,225^\circ+n180^\circ, n\in Z\}$$ However, as we notice, $45^\circ+n180^\circ$ and $225^\circ+n180^\circ$ represents the same values, so we would only pick one of them. Here I would go with $45^\circ+n180^\circ$ $$2\theta=\{45^\circ+n180^\circ, n\in Z\}$$ Thus, $$\theta=\{22.5^\circ+n90^\circ,n\in Z\}$$