Question

How do you solve $\cos 2x - \sin x = 1$in the interval $[0,360]?$

Answer 1

Hint: Here we will use the trigonometric identity and will simplify the equation using it. Also, will use the All STC law to find the angles in the interval between $0^\circ $to $360^\circ $.

Complete step-by-step solution:
Here we will use the identity for $\cos 2x = 1 - 2{\sin ^2}x$
Place the above value in the given equation.
$1 - 2{\sin ^2}x - \sin x = 1$
Like terms with the same value and same sign on the opposite sides of the equations cancel each other.
$ - 2{\sin ^2}x - \sin x = 0$
Take the negative sign common from the above equation.
$2{\sin ^2}x + \sin x = 0$
Take the common factor from the above equation from both the terms in it.
$\sin x(2\sin x + 1) = 0$
Hence, the roots of the equation will be either
$\sin x = 0$ or $2\sin x + 1 = 0$
First solve for $\sin x = 0$
Referring to the trigonometric table for values,
$ \Rightarrow x = 0^\circ ,180^\circ $ ….. (A)
Now, $2\sin x + 1 = 0$
Take constant on the right hand side of the equation, when you move any term from one side of the equation to other then the sign of the term also changes. Positive terms become negative and vice versa.
$2\sin x = - 1$
Term multiplicative on one side if moved to the opposite side then it goes to the denominator.
$\sin x = - \frac{1}{2}$
Using All STC rule, sine function is negative in the third and the fourth quadrant.
$ \Rightarrow x = 210^\circ ,330^\circ $ ….. (B)
Hence, the resultant answer is

$x = 0^\circ ,180^\circ $
or
$x = 210^\circ ,330^\circ $

Note: Remember the All STC rule, it is also known as 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 $ ).