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.