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# How do you solve $\sin 4x + \sin 2x = 0$ using the product and sum formulas?

Last updated date: 16th Sep 2024
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Hint: Here we will use the sine angle identity will replace the formula and find the factors for the given equation. Then we will use the All STC rule to find the required resultant angles.

Complete step-by-step solution:
Take the given equation: $\sin 4x + \sin 2x = 0$
The above equation can be re-written as: $\sin 2(2x) + \sin 2x = 0$
Using the identity: $\sin 2\theta = 2\sin \theta \cos \theta$ in the above equation.
$\Rightarrow 2\sin 2x\cos 2x + \sin 2x = 0$
Take common multiples from the above equation.
$\Rightarrow \sin 2x(2\cos 2x + 1) = 0$
From the above equation there are two cases:
Case (I)
$\Rightarrow \sin 2x = 0$
Take the inverse of sine on both the sides of the above equation.
$\Rightarrow {\sin ^{ - 1}}(\sin 2x) = {\sin ^{ - 1}}(0)$
Referring to the trigonometric table for sine value,
$\Rightarrow {\sin ^{ - 1}}(\sin 2x) = {\sin ^{ - 1}}(\sin 0^\circ )$
Sine inverse and sine cancel each other.
$\Rightarrow 2x = 0$
Make “x” has the subject
$\Rightarrow x = \dfrac{0}{2}$
Zero upon anything is always zero.
$\Rightarrow x = 0$
We know that in a circle for every $360^\circ$ angle all values are repeated.
$\therefore x = 0,2\pi ,4\pi ,.....$ ….(A)
Case (II)
$2\cos 2x + 1 = 0$
Make the subject angle and move other terms on the opposite side. Remember when you move any term from one to another, the sign of the term also changes. Positive terms become negative.
$\Rightarrow 2\cos 2x = - 1$
Term multiplicative on one side if moved to the opposite side, then it goes to the denominator.
$\Rightarrow \cos 2x = - \dfrac{1}{2}$
Cosine is negative in the second and third quadrant.
$\Rightarrow 2x = \dfrac{{2\pi }}{3},\dfrac{{4\pi }}{3},\dfrac{{8\pi }}{3},.....$
Simplify the above equation,
$\Rightarrow x = \dfrac{\pi }{3},\dfrac{{2\pi }}{3},\dfrac{{4\pi }}{3},.....$ ….(B)
Equations (A) and (B) are the required solution.

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