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Question

Question: How do you solve \(4\sin x=3\) for \({{0}^{\circ }} < x < {{360}^{\circ }}\) ?...

How do you solve 4sinx=34\sin x=3 for 0<x<360{{0}^{\circ }} < x < {{360}^{\circ }} ?

Explanation

Solution

In this question we have to find the values of x lying in the interval 0<x<360{{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 4sinx=34\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<x<360{{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 4sinx=34\sin x=3 …. (i)
From equation (i) we get that sinx=34\sin x=\dfrac{3}{4}.
This means that x=sin1(34)x={{\sin }^{-1}}\left( \dfrac{3}{4} \right).
If we simplify this we get that x=48.6x={{48.6}^{\circ }}.
We also know that sinx=sin(180x)\sin x=\sin ({{180}^{\circ }}-x), where x is in degrees.
With this we get that ,
sin(48.6)=sin(18048.6)=sin(131.4)\sin ({{48.6}^{\circ }})=\sin ({{180}^{\circ }}-{{48.6}^{\circ }})=\sin ({{131.4}^{\circ }})

However, it is given that sinx=34\sin x=\dfrac{3}{4}.
This means that sin(131.4)=34\sin ({{131.4}^{\circ }})=\dfrac{3}{4}.
Therefore, x=131.4x={{131.4}^{\circ }}
And since x=48.6x={{48.6}^{\circ }} and x=131.4x={{131.4}^{\circ }} between 0{{0}^{\circ }} and 360{{360}^{\circ }}, the correct solutions of the given equation are x=48.6x={{48.6}^{\circ }} and x=131.4x={{131.4}^{\circ }}

Hence, the correct solutions of the given equation are x=48.6x={{48.6}^{\circ }} and x=131.4x={{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 2π2\pi radians.Since they are periodic, they repeat themselves after every interval of 2π2\pi radians. Therefore, note that trigonometric equations have more than one solution.