Question

How do you solve $4\sin x=3$ for ${{0}^{\circ }} < x < {{360}^{\circ }}$ ?

Answer 1

Hint: In this question we have to find the values of x lying in the interval ${{0}^{\circ }} < x <{{360}^{\circ }}$, that satisfy the given equation. Use inverse trigonometric functions to find the solution of the equation. It may have more than one solution.

Complete step by step answer:
The equation given in the question is $4\sin x=3$ and we are asked to solve the equation for the given interval of x. This means that we have to find the values of x lying in the interval ${{0}^{\circ }} < x < {{360}^{\circ }}$, that satisfy the given equation. We can see that the equation consists of a sine function (i.e. sinx). Let us first find all the values of x that satisfy the given equation.

The equation says that $4\sin x=3$ …. (i)
From equation (i) we get that $\sin x=\dfrac{3}{4}$.
This means that $x={{\sin }^{-1}}\left( \dfrac{3}{4} \right)$.
If we simplify this we get that $x={{48.6}^{\circ }}$.
We also know that $\sin x=\sin ({{180}^{\circ }}-x)$, where x is in degrees.
With this we get that ,
$\sin ({{48.6}^{\circ }})=\sin ({{180}^{\circ }}-{{48.6}^{\circ }})=\sin ({{131.4}^{\circ }})$

However, it is given that $\sin x=\dfrac{3}{4}$.
This means that $\sin ({{131.4}^{\circ }})=\dfrac{3}{4}$.
Therefore, $x={{131.4}^{\circ }}$
And since $x={{48.6}^{\circ }}$ and $x={{131.4}^{\circ }}$ between ${{0}^{\circ }}$ and ${{360}^{\circ }}$, the correct solutions of the given equation are $x={{48.6}^{\circ }}$ and $x={{131.4}^{\circ }}$

Hence, the correct solutions of the given equation are $x={{48.6}^{\circ }}$ and $x={{131.4}^{\circ }}$.

Note: Note that all the trigonometric functions like sine and cosine functions are periodic functions. The period of trigonometric functions is equal to $2\pi $ radians.Since they are periodic, they repeat themselves after every interval of $2\pi $ radians. Therefore, note that trigonometric equations have more than one solution.