Solving a linear trigonometric equation in double angle form

Key takeaway: The equation $4\sin 2\theta + 3\cos 2\theta = 0$ becomes manageable once we rewrite it as a tangent relationship for $2\theta$.

Recognize the structure

The equation $4\sin 2\theta + 3\cos 2\theta = 0$ is a linear trigonometric equation in the double angle $2\theta$. The two trig terms share the same angle, so the best move is to isolate one term and divide by the other when it is safe to do so.

Here, the goal is to convert the sum into a ratio. Since both sine and cosine appear with coefficients, this is a standard tangent-type equation rather than a quadratic identity problem. The interval for $\theta$ is $0^\circ \le \theta < 360^\circ$, so the corresponding interval for $2\theta$ will run from $0^\circ$ to $720^\circ$.

Use a tangent identity

Start from

$4\sin 2\theta + 3\cos 2\theta = 0$.

Move one term to the other side:

$4\sin 2\theta = -3\cos 2\theta$.

If $\cos 2\theta \ne 0$, divide both sides by $\cos 2\theta$:

$4\tan 2\theta = -3$,

so

$\tan 2\theta = -\frac{3}{4}$.

Now find the reference angle. Since $\tan^{-1}(3/4) \approx 36.9^\circ$, the angle for tangent is about $36.9^\circ$ away from each axis. Tangent is negative in Quadrants II and IV, so for $2\theta$ in $[0^\circ,720^\circ)$ the solutions are

$2\theta \approx 143.1^\circ, 323.1^\circ, 503.1^\circ, 683.1^\circ$.

Convert back to $\theta$ and check the range

Divide each angle by 2:

$\theta \approx 71.6^\circ, 161.6^\circ, 251.6^\circ, 341.6^\circ$.

These all lie in the required interval $0^\circ \le \theta < 360^\circ$, so they are the complete set of solutions.

The matching answer choice is the list above, which confirms the tangent-based approach is correct.

Common mistake to avoid

A frequent error with $4\sin 2\theta + 3\cos 2\theta = 0$ is to stop after solving only one cycle of $2\theta$. Because the angle is doubled, the equation has twice as many solutions for $\theta$ within the original interval.

Another mistake is dividing by $\sin 2\theta$ instead of $\cos 2\theta$ without checking whether that would exclude valid solutions. The tangent form is easiest here, but only if you remember to cover the full interval for $2\theta$ before halving back to $\theta$.

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Pitfalls the pros know 👇 The first place this problem can go wrong is when you solve $\tan 2\theta = -3/4$ but only list two angles for $2\theta$ instead of four. Because $\theta$ ranges from $0^\circ$ to $360^\circ$, the doubled angle $2\theta$ ranges over two full turns, so every tangent solution repeats once more after adding $180^\circ$. Another subtle trap is forgetting that tangent is negative in Quadrants II and IV, which leads to the wrong reference-angle placement. A final issue is arithmetic: after finding $2\theta \approx 143.1^\circ$ etc., you must divide by 2 at the end and keep the answers in the original interval. If you divide too early or round too aggressively, the listed choices can look slightly off even when the method is right.

What if the problem changes? If the equation were changed to $4\sin 2\theta - 3\cos 2\theta = 0$ on the same interval $0^\circ \le \theta <360^\circ$, the setup would be almost identical, but the tangent ratio would become $\tan 2\theta = 3/4$ instead of a negative value. That changes the quadrants that contain solutions: tangent would be positive in Quadrants I and III for each $180^\circ$ cycle. The corresponding $2\theta$ values would shift by the same reference angle, and after dividing by 2 you would get a different set of four angles. This variant is useful because it tests whether you understand that the sign in front of the cosine term controls the sign of tangent, while the angle-doubling still forces you to account for the entire $0^\circ$ to $720^\circ$ range before converting back to $\theta$.

Tags: double angle identity, tangent form equation, trigonometric interval solutions