To solve the given equation, we will first factor it and use the Zero Product Property to solve for sinθ. Then we will use the unit circle to find the exact values of θ that satisfy the equation.
Solving the Equation for sinθ
Let's start by rewriting the equation so that all of the terms are on the left-hand side, so we can factor it.
We obtained two values for sin θ. The sine of an angle in standard position is the y-coordinate of the point of intersection P of its terminal side and the unit circle.
P(x,y)=(cosθ,sinθ)
To solve the first equation, sin θ=0, we need to consider the points on the unit circle that have a y-coordinate of 0.
We found two solutions to the equation sin θ=0. These are also two of the solutions to the given equation.
θ=0 radians and θ=π radians
To solve the second equation, sin θ =- 12, we need to consider the points on the unit circle that have a y-coordinate of - 12. Recall that sine is negative in Quadrants III and IV.
We can now draw two congruent right triangles, each with a leg on the x-axis. Since the radius of the unit circle is 1, the hypotenuse of each triangle is also 1. Furthermore, since the y-coordinate of both points is - 12, the length of the leg which is not on the x-axis of each triangle is 12.
For both right triangles, one of the legs is half the hypotenuse, so they are 30^(∘)-60^(∘)-90^(∘) triangles. In this type of triangle, the smaller angle measures 30^(∘) or π6 radians. With this information and knowing that half a turn measures π radians and a full turn measures 2π radians, we can calculate the desired angle measures. We will add π6 to π, and subtract π6 from 2π.
We found two solutions to the equation sin θ=- 12. These are also the solutions to the given equation.
θ= 7π6 radians and θ= 11π6 radians
After all, the given equation has four solutions.
0 radians, π radians, 7π6 radians, 11π6 radians
