Solving a shifted sine equation with triple angle

Key concept: The equation $2 + 2\sin 3\varphi = 1$ simplifies quickly once $\sin 3\varphi$ is isolated and its exact values are found.

Step by step

Isolate the trigonometric function

The equation $2 + 2\sin 3\varphi = 1$ is already set up for isolation. Subtract 2 from both sides to get

$2\sin 3\varphi = -1$.

Then divide by 2:

$\sin 3\varphi = -\frac{1}{2}$.

This is now a standard sine equation. The variable is $3\varphi$, so the angle inside the sine function must be solved first, and only afterward do we divide by 3.

Find all angles for $3\varphi$

Since $\sin x = -\frac{1}{2}$ at angles with reference angle $30^\circ$, the solutions for $x = 3\varphi$ are

$3\varphi = 210^\circ, 330^\circ, 570^\circ, 690^\circ$,

because $0^\circ \le \varphi < 180^\circ$ implies

$0^\circ \le 3\varphi < 540^\circ$.

Notice that only the angles inside that range are allowed. The values $210^\circ$ and $330^\circ$ come from the first revolution, and $570^\circ$ and $690^\circ$ are not actually in the interval $[0^\circ,540^\circ)$, so they must be rejected.

That leaves only

$3\varphi = 210^\circ, 330^\circ$.

Convert back to $\varphi$

Now divide by 3:

$\varphi = 70^\circ, 110^\circ$.

These are the complete solutions in the range $0^\circ \le \varphi < 180^\circ$.

A quick check confirms them: $3(70^\circ)=210^\circ$ and $\sin 210^\circ = -1/2$, while $3(110^\circ)=330^\circ$ and $\sin 330^\circ = -1/2$. Substituting back gives $2 + 2(-1/2)=1$, which matches the equation exactly.

Why this method works

The structure $2 + 2\sin 3\varphi = 1$ is best handled by isolating the sine term first, rather than trying to use a more complicated identity. Once the sine value is known, the solution set comes from standard unit-circle values.

Because the angle is tripled, the valid $3\varphi$ values must be filtered using the interval induced by the original domain. That interval check is what prevents extra answers from slipping in.

Pitfall alert

The first trap in $2 + 2\sin 3\varphi = 1$ is solving $\sin 3\varphi = -1/2$ and immediately listing every sine angle you know, without checking whether $3\varphi$ stays inside the allowed range. Since $0^\circ \le \varphi < 180^\circ$, the triple angle only runs from $0^\circ$ to $540^\circ$, so any angle above that must be rejected. Another common error is dividing by 3 too early and then trying to apply the sine pattern to $\varphi$ itself; the reference angle belongs to $3\varphi$, not to $\varphi$. Finally, some students forget that sine is negative in Quadrants III and IV, which can lead to the wrong pair of angles. The safest workflow is isolate, solve for the inside angle, filter by interval, and only then divide by 3.

Try different conditions

If the equation were changed to $2 + 2\sin 3\varphi = 3$ on the same interval $0^\circ \le \varphi <180^\circ$, the isolation step would give $\sin 3\varphi = \frac{1}{2}$ instead of a negative value. That changes the unit-circle angles for $3\varphi$ to $30^\circ$ and $150^\circ$, along with any additional coterminal angles that still fit inside $[0^\circ,540^\circ)$. After that, you would divide each valid $3\varphi$ value by 3 to get the corresponding $\varphi$ values. This variant checks whether you can keep the same algebraic process while adapting the sign and the quadrants correctly. It also tests whether you remember that the interval restriction applies to the inside angle first, not just the final answer.

Further reading

triple angle sine, unit circle values, restricted interval solving