Question

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

Answer 1

Hint: The trigonometric functions are the circular functions with the function of an angle and are related to the angles of a triangle with the lengths of its sides. In the Cartesian coordinate system, the circle centred origin $ o(0,0) $ is the unit circle, where the points distance from the origin is always one. The x-coordinate to the point of intersection is equal to the $ \cos \theta $ . Here, we will use the properties of the cosine functions and the All STC rule.

Complete step-by-step answer:
Since, cosine is an even function.
By All STC rules, cosine is negative in the second quadrant.
Now, as per the quadrant division the given angle should lie between \[\dfrac{\pi }{2}{\text{ to}}\pi \]
 $ \cos (\pi - x) = - \cos x $
Also, we know that $ \cos \dfrac{\pi }{3} = \dfrac{1}{2} $
 $ \Rightarrow \theta = \pi - \dfrac{\pi }{3} $
Take LCM on the right hand side of the equation and simplify-
 $ \Rightarrow \theta = \dfrac{{2\pi }}{3} $
Hence, the required general solution for $ \cos \theta = \dfrac{{ - 1}}{2} $ is $ \theta = \dfrac{{2\pi }}{3} $

Note: Remember the All STC rule, it is also known as the ASTC rule in geometry. It states that all the trigonometric ratios in the first quadrant ( $ 0^\circ \;{\text{to 90}}^\circ $ ) are positive, sine and cosec are positive in the second quadrant ( $ 90^\circ {\text{ to 180}}^\circ $ ), tan and cot are positive in the third quadrant ( $ 180^\circ \;{\text{to 270}}^\circ $ ) and sin and cosec are positive in the fourth quadrant ( $ 270^\circ {\text{ to 360}}^\circ $ ). Remember the trigonometric values for all the functions for the angles $ 0^\circ ,30^\circ ,45^\circ ,60^\circ ,90^\circ $ for direct substitutions.