Question

How do you solve $\sin (2x)\cos (x) = \sin (x)$ ?

Answer 1

Hint : First of all take all the terms on one side of the equation and use different trigonometric identities and simplify for the resultant value for “x” by using general solution formulae of sin and cos.

Complete step by step solution:
Take the given expression: $\sin (2x)\cos (x) = \sin (x)$
Move term from right hand side to the left hand side. When you move any term from one side to another, the sign of the term also changes. Positive terms become negative and vice versa.
$\sin (2x)\cos (x) - \sin (x) = 0$
Use the identity $\sin 2x = 2\sin x\cos x$ and place it in the above expression.
$2\sin (x)\cos (x)\cos (x) - \sin (x) = 0$
Take the common multiple from both the terms in the above expression.
$\Rightarrow \sin x(2{\cos ^2}x - 1) = 0$
The above expression suggests –
$\sin x = 0$ …. (A)
or $2{\cos ^2}x - 1 = 0$
Simplify the above expression –
$2{\cos ^2}x = 1$
Term multiplicative on one side if moved to the opposite side then it goes to the denominator.
${\cos ^2}x = \dfrac{1}{2}$
Take square-root on both the sides of the above expression –
$\cos x = \pm \dfrac{1}{{\sqrt 2 }}$ ….(B)
Referring the All STC rule, for the expression for the equations (A) and (B)
$\Rightarrow x = n\pi$ or $x = 2n\pi \pm \dfrac{\pi }{4}$
This is the required solution.
So, the correct answer is “ $x = n\pi$ or $x = 2n\pi \pm \dfrac{\pi }{4}$ ”.

Note: Remember the trigonometric table having different angles of measures. 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 ( $270^\circ {\text{ to 360}}^\circ$ ).