Question

How to solve this? $\sin x + \sin 2x + \sin 3x = 0$

Answer 1

Hint : Here we will arrange the three given terms in two plus one and then will use a formula for the sum of two sine functions. Also, then by using the trigonometric identity and by using the All STC rule then will find the resultant required value.

Complete step by step solution:
Take the given equation: $\sin x + \sin 2x + \sin 3x = 0$
Rearranging the terms, shifting the second term with the third term and vice-versa.
$\sin x + \sin 3x + \sin 2x = 0$
Make the pair of first two terms,
$\underline {\sin x + \sin 3x} + \sin 2x = 0$
Use the trigonometric identity: $\sin a + \sin b = 2\sin \left( {\dfrac{{a + b}}{2}} \right).\cos \left( {\dfrac{{a - b}}{2}} \right)$ in the first paired terms.
$\Rightarrow 2\sin \left( {\dfrac{{3x + x}}{2}} \right).\cos \left( {\dfrac{{3x - x}}{2}} \right) + \sin 2x = 0$
Simplify the above equation-
$\Rightarrow 2\sin \left( {2x} \right).\cos \left( x \right) + \sin 2x = 0$
Take common term from the above equation –
$\Rightarrow \sin \left( {2x} \right)(2\cos \left( x \right) + 1) = 0$
To get the values for “x”, either one of the above two factors must be equal to zero.
$\sin 2x = 0$ or $2\cos x + 1 = 0$
Make required “x” the subject. When you move any term from one side to another, the sign of the term changes. Positive term becomes the negative and vice-versa. Also, the term multiplicative on one side if moved to the opposite side then it goes to the denominator.
$\Rightarrow 2x = {\sin ^{ - 1}}\left( {\dfrac{0}{2}} \right)$
or
$2\cos x = - 1 \\\ \Rightarrow \cos x = - \dfrac{1}{2} \\\ \Rightarrow x = {\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) \;$
Referring to the trigonometric table for values and All STC rule-
$x = \dfrac{{k\pi }}{2}$ or $x = \pm \dfrac{{2\pi }}{3}$
Hence, the required solution is $x = \dfrac{{k\pi }}{2}$ or $x = \pm \dfrac{{2\pi }}{3}$
So, the correct answer is “ $x = \dfrac{{k\pi }}{2}$ or $ x = \pm \dfrac{{2\pi }}{3} ”.

Note : Also 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 ( $270^\circ {\text{ to 360}}^\circ$ ).