Question

The principle solution of $\cos \theta = \dfrac{{ - 1}}{2}$is
$
  {\text{A) }}\dfrac{{2\pi }}{3} \\
  {\text{B) }}\dfrac{\pi }{6} \\
  {\text{C)}}\dfrac{{4\pi }}{3} \\
  {\text{D)}}\dfrac{{7\pi }}{6} \\
 $

Answer 1

Hint: Start the solution by defining what a principle solution means in trigonometry. After this according to the definition of the principle solution try to find the value of $\theta $ when $\cos \theta = \dfrac{{ - 1}}{2}$ where the value of $\theta $must lie in the interval of 0 and $2\pi $. This may have more than one value as mentioned in the question.

Complete step by step answer:
we will start the solution by noting down what the principle solution means.
We already know that the values of $\cos x$ repeat after the interval of $2\pi $$\pi = \dfrac{{4\pi }}{3}$
Equations involving the variable $0 \leqslant x \leqslant 2\pi $, their solutions are called principal
Hence from the above definition we have understood that we have to find the value of $\theta $when $\cos \theta = \dfrac{{ - 1}}{2}$ in the interval 0 and $2\pi $.
We all know that $\cos \theta = \dfrac{{ - 1}}{2}$ when $\theta = \dfrac{{2\pi }}{3}$ and $\theta = \dfrac{{4\pi }}{3}$
Hence this are our principle solutions for $\cos \theta = \dfrac{{ - 1}}{2}$

Hence the principle solutions for $\cos \theta = \dfrac{{ - 1}}{2}$ are (A) $\dfrac{{2\pi }}{3}$ and (C) $\dfrac{{4\pi }}{3}$

Note: questions on determining the principal value can be asked for different trigonometric equations. The steps of solving are the same as used above. But the only changes that are supposed to be remembered are that different trigonometric functions have different values in the same given interval.