Question

How do you solve $\sin 4x - 2\sin 2x = 0$ in the interval $\left[ {0,360} \right]$?

Answer 1

Hint: Given the trigonometric expression. We have to find the value of x that must lie between zero and $360$. First, we will apply the trigonometric identities to the expression. Then, simplify the expression by taking out the common terms from the bracket. Then, find the value of x by setting each expression equal to zero. Then, we will apply the general solution of the equation.

Formula used:
The formula for $\sin \left( {A + B} \right)$ is given as:
$\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$
The general solution of the equation, $\sin x = 0$ is given as:
$x = n\pi $
The general solution of the equation, $\cos \theta = \cos \alpha $ is given as:
$\theta = 2n\pi \pm \alpha $

Complete step-by-step answer:
Given expression is $\sin 4x - 2\sin 2x = 0$
First, we will rewrite the expression $\sin 4x$as a sum of $2x$ and $2x$.
$ \Rightarrow \sin \left( {2x + 2x} \right) - 2\sin 2x = 0$
Now, we will apply the sum formula to the expression.
$ \Rightarrow \sin 2x\cos 2x + \cos 2x\sin 2x - 2\sin 2x = 0$
On combining like terms, we get:
$ \Rightarrow 2\sin 2x\cos 2x - 2\sin 2x = 0$
Now, take out the common terms.
$ \Rightarrow 2\sin 2x\left( {\cos 2x - 1} \right) = 0$
Now, set each factor equal to zero.
$ \Rightarrow 2\sin 2x = 0{\text{ or }}\cos 2x - 1 = 0$
$ \Rightarrow \sin 2x = 0{\text{ or }}\cos 2x = 1$
Now, apply the general solution to the expression $\sin 2x = 0$, we get:
$ \Rightarrow 2x = 0,\pi ,2\pi ,3\pi ,4\pi $
Now, divide both sides by $2$.
$ \Rightarrow x = 0,\dfrac{\pi }{2},\dfrac{{2\pi }}{2},\dfrac{{3\pi }}{2},\dfrac{{4\pi }}{2}, \ldots $
$ \Rightarrow x = 0,\dfrac{\pi }{2},\pi ,\dfrac{{3\pi }}{2},2\pi \ldots $
Now, substitute $180^\circ $ for $\pi $ into the expression.
$ \Rightarrow x = 0,90^\circ ,180^\circ ,270^\circ ,360^\circ , \ldots $ …… (1)
First, rewrite the expression $\cos 2x = 1$, by writing $1$ in terms of $\cos \theta $.
$\cos 2x = \cos 0^\circ $
Now we will apply the general solution to the expression $\cos 2x = \cos 0^\circ $.
$ \Rightarrow 2x = 0,2\pi ,4\pi ,8\pi , \ldots $
Now, divide both sides by $2$.
 $ \Rightarrow x = 0,\dfrac{{2\pi }}{2},\dfrac{{4\pi }}{2},\dfrac{{8\pi }}{2}, \ldots $
$ \Rightarrow x = 0,\pi ,2\pi , \ldots $
Now, substitute $180^\circ $ for $\pi $ into the expression.
$ \Rightarrow x = 0,180^\circ ,360^\circ $ …... (2)
Now, combine the solutions from equation (1) and (2) in the interval $\left[ {0,360} \right]$.
$ \Rightarrow x = 0,90^\circ ,180^\circ ,270^\circ ,360^\circ $

Final answer: Hence the solution of the expression is $0,90^\circ ,180^\circ ,270^\circ ,360^\circ $

Note:
In such types of questions students mainly get confused in applying the formula. As they don't know which formula they have to apply. So, when the trigonometric function is given, then the student must apply the appropriate trigonometric formula and substitute the values into the formula.