Question

How do you solve \[\sin 3x=\dfrac{1}{2}\]?

Answer 1

Hint: In this problem, we have to solve for x in the given trigonometric expression. We should know the sin values to solve this problem. We know that the value of sin 30 degrees is \[\dfrac{1}{2}\], so we have to equate it and cancel the sin terms to get the value of x. We also know that the trigonometric unit circle gives another arc x that has the same sin value.

Complete step by step answer:
We know that the given trigonometric expression is,
\[\sin 3x=\dfrac{1}{2}\] ……. (1)
We also know that sin 30 degree is equal to \[\dfrac{1}{2}\], we can write the expression (1) as
\[\sin 3x=\sin \dfrac{\pi }{6}\] ……. (2)
Now we can cancel the sin terms, we get
\[\Rightarrow 3x=\dfrac{\pi }{6}\]
Now, we can divide 3 on both sides, we get
\[\Rightarrow x=\dfrac{\pi }{18}\]
We also know that the trigonometric unit circle gives another arc x that has the same sin value.
That is;
\[\begin{align}
  & \Rightarrow \sin 3x=\sin \left( \pi -\dfrac{\pi }{6} \right) \\
 & \Rightarrow \sin 3x=\sin \dfrac{5\pi }{6} \\
\end{align}\]
Now we can cancel the sin terms on both sides, we get
\[\Rightarrow 3x=\dfrac{5\pi }{6}\]
Now, we can divide 3 on both sides we get
\[\Rightarrow x=\dfrac{5\pi }{18}\]
We know that the period of \[\sin 3x\] function is \[\dfrac{2\pi }{3}\] so value will repeat every \[\dfrac{2\pi }{3}\] radians in both direction.
Therefore, x = \[\dfrac{\pi }{18}+\dfrac{2\pi n}{3},\dfrac{5\pi }{18}+\dfrac{2\pi n}{3}\] for any integer n.

Note:
Students make mistakes in the sin values for degrees, we should always remember basic sin and cos values to solve these types of problems. We should also remember that the trigonometric unit circle gives another arc x that has the same sin value.