Question

Find the principal solutions of $\cos x = - \dfrac{1}{2}$ .

Answer 1

Hint : In this question, we have been given to find the principal solutions. Students get confused when the word ‘principal’ comes. They don’t know in what range they have to find the answer. But, one thing to always remember is that for calculating principal solutions of any trigonometric expression, we find the value of the argument or the value of the angle (here $x$ ) in the range between 0 and $2\pi$ , i.e., in the range between $0^\circ$ and $360^\circ$ , or we can say a full circle. A thing to remember is that there might be more than one answer for the trigonometric expression in the question, so the students do not need to worry if they come across something like that.

Complete step-by-step answer :
In this question, we need to find the principal solutions – the solutions which lie in the range $\left[ {0,2\pi } \right]$ .
Now, $\cos \left( {\pi - x} \right) = - \cos x$
and also, $\cos \left( {\pi + x} \right) = - \cos x$
Now, $\cos \dfrac{\pi }{3} = \dfrac{1}{2}$
So, $\cos \left( {\pi - \dfrac{\pi }{3}} \right) = \cos \dfrac{{2\pi }}{3} = - \dfrac{1}{2}$
Similarly, $\cos \left( {\pi + \dfrac{\pi }{3}} \right) = \cos \dfrac{{4\pi }}{3} = - \dfrac{1}{2}$
Hence, the principal solution of $\cos x = - \dfrac{1}{2}$ is $x = \dfrac{{2\pi }}{3},\dfrac{{4\pi }}{3}$ .
So, the correct answer is “ $\cos x = - \dfrac{1}{2}$ is $x = \dfrac{{2\pi }}{3},\dfrac{{4\pi }}{3}$ .”.

Note : So, we saw that in solving questions like these, when it has been given that we need to find the solution of the given trigonometric expression in the question in the range between 0 and $2\pi$ , or $0^\circ$ and $360^\circ$ , or we can say a full circle. For the trigonometric expressions, the solution, i.e., the value of the argument is the value of the angle. A thing to note is that while solving these trigonometric expressions given in the question, we might come across two principal solutions in the range, and so, we need not worry about it.