Question

How do you solve the equation $\sin \theta = 0.91$ for $0 < \theta < 2\pi ?$

Answer 1

Hint: To determine the solution of the given equation in the given interval, one should know that the sine function gives positive values in the first and second quadrant, so you have to find the solution of the given equation in both the quadrants. And in order to find the solution in both the quadrants, take the inverse function of sine and then solve accordingly for both the quadrants.

Complete step by step solution:
 In order to solve the given trigonometric equation $\sin \theta = 0.91$ in the interval $0 < \theta < 2\pi $ we will take the inverse sine function in both sides of the equation, we will get
$
   \Rightarrow {\sin ^{ - 1}}\left( {\sin \theta } \right) = {\sin ^{ - 1}}\left( {0.91} \right) \\
   \Rightarrow \theta = {\sin ^{ - 1}}\left( {0.91} \right) \\
 $
Since we know that the sine function gives positive values in two quadrants, therefore we will get two solutions for the given equation and the solutions can be given as
Solution for the first quadrant is given as
$
   \Rightarrow \theta = {\sin ^{ - 1}}\left( {0.91} \right) \\
   \Rightarrow \theta = {65.1^0} \\
 $
And solution for the second quadrant is given as
$
   \Rightarrow \theta = {180^0} - {65.1^0} \\
   \Rightarrow \theta = {114.9^0} \\
 $

Therefore $\theta = {65.1^0}\;{\text{and}}\;{114.9^0}$ is the required solution for the given equation in the interval $0 < \theta < 2\pi $

Note: We have found the answer in degrees, if the required answer the question wants is in some different unit of angle like radian (which is standard unit of angle measurement), convert the answer into that unit by the conversion formula of the two units. Also the solution we have calculated is a particular solution which means that the solution is lying in between the given interval whereas there are two more types of solutions that are known as principle solution and general solution.