Solve tan and sin trigonometric equations for theta

Key takeaway: This is one of those trigonometry questions where the algebra does most of the work. Once the equation is rearranged, factoring reveals the allowed angles, and then the domain or marking scheme usually trims the list down.

Start with $\tan\theta+2\sin\theta=3\tan\theta-6\sin\theta$ Move everything to one side: $2\tan\theta-8\sin\theta=0$ Now write $\tan\theta$ as $\dfrac{\sin\theta}{\cos\theta}$: $2\frac{\sin\theta}{\cos\theta}-8\sin\theta=0$ Multiply through by $\cos\theta$: $2\sin\theta-8\sin\theta\cos\theta=0$ Factor out $2\sin\theta$: $2\sin\theta(1-4\cos\theta)=0$ So either $\sin\theta=0$ or $1-4\cos\theta=0 \Rightarrow \cos\theta=\frac14.$

In the given solution, the valid angle is $\theta\approx 75.5^\circ.$ If the domain excludes the angles where $\sin\theta=0$ or where $\tan\theta$ is undefined, then that may be the only accepted answer.

So the key algebraic step is the factorisation $2\sin\theta(1-4\cos\theta)=0.$

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Pitfalls the pros know 👇 A frequent mistake is forgetting that converting $\tan\theta$ to $\sin\theta/\cos\theta$ introduces a denominator, so you must keep an eye on where $\cos\theta=0$. Another common error is stopping at $\cos\theta=\frac14$ and missing the other factor $\sin\theta=0$.

What if the problem changes? If the equation were solved over $0^\circ\le \theta<360^\circ$, then $\sin\theta=0$ would give additional angles such as $0^\circ,180^\circ,360^\circ$ depending on the interval. The $\cos\theta=\frac14$ solution would also produce more than one angle in a full-turn domain.

Tags: factorisation, trigonometric identity, principal angle