Question

Find the general solution of $\cos \theta = - \dfrac{1}{2}$

Answer 1

To find the general solution of any trigonometric function. We first write the given value in trigonometric function form and then add periodic cycles of respective trigonometric functions.

Formulas Used: If $\cos \theta = \cos \alpha$ which implies $\theta = 2n\pi \pm (\alpha )$ where n is any integer.

Complete step-by-step solution:
Here, the given trigonometric function is$\cos \theta = - \dfrac{1}{2}$. …………………….(i)
We know that $\cos \theta$ is negative in either 2nd quadrant (${90^0} < \theta < {180^0}$) or in 3rd quadrant$\left( {{{180}^0} < \theta < {{270}^0}} \right)$.
Also, we know that the value of$\cos {120^0} = - \dfrac{1}{2}$. …………………….(ii)
From above (i) and (ii) equations we have right hand side equal,
$\therefore$ $\cos \theta = \cos {120^0}$ Or
$\cos \theta = \cos \left( {\dfrac{{2\pi }}{3}} \right)$ $\left( {\because \dfrac{{2\pi }}{3} = {{120}^0}} \right)$
Also, we know that $\cos \theta$ is a periodic function having a period of$2\pi$. Therefore in general its periodic cycle is written as$2n\pi$.
Hence, if $\cos \theta = \cos \alpha$then we can write:
$\theta = 2n\pi \pm \alpha$
Where $\alpha$ is$\dfrac{{2\pi }}{3}$.
Therefore, from above we have $\theta = 2n\pi \pm \left( {\dfrac{{2\pi }}{3}} \right)$
Which is the required general solution of$\cos \theta = - \dfrac{1}{2}$.
Hence, from above we see that general solution of $\cos \theta = - \dfrac{1}{2}$ is $\theta = 2n\pi \pm \left( {\dfrac{{2\pi }}{3}} \right)$

Note: In trigonometry we know that trigonometric functions are periodic functions. Therefore, every function has its own general formulas. Here $\cos \theta$ is a periodic function therefore it gives the same values but for different periodic cycles.