# Trigonometric Equations

Example 1

Find all the angles for which

Viewed in a unit circle, sine is the y-value, so at two angles the sine value is 0,5

The first angle at which the sine has a value of 0,5 is π / 6, so

The period of the sine is 2π, so every full circle the sine has the same value. All solutions to the equation sin (x) = 0,5 are

The previous n is an integer. It can be positive or negative, meaning we can rotate the circle in either direction.

Example 2

Solve the equation

Trigonometric equations are treated like any other equation. The first step is to get the equation to sin (x) = a

From the table we get the first angle from which the equation solves. It is x = π / 3 and the other angle is x = π-π / 3 = 2π / 3, so the solutions to the equation are

Example 3

Solve the equation

The angles at which the sine have these values are π / 4 + n2π and 3π / 4 + n2π. Then we get

When both equations are divided by two, we can solve x

where n is an integer.

When solving the equation, it is good to note that the period needs to be divided also. Just like in any equation, all terms are divided.

Example 4

Solve the equation

First modify the equation to the form cos (x) = a

With a calculator, the angle at which the cosine is 1/3 is 1,2309 .... The solutions of the equation are

where n is an integer.

Example 5

Solve the equations

in a, the angles must be equal or supplementary.

From these equations we can solve the angle x. Move x to the left of the equations.

n is an integer and contains both negative and positive integers, so we can give the first answer without a minus sign.

The equation in b can be modified to

The tangent is not defined by the angles π / 2 + nπ, where n is an integer. That is, at the angles where the cosine has a value of 0. The period of the tangent is π, so the solution of the equation is

where n is an integer.

Example 6

Solve the equation

Modify the equation so that there is only sine or cosine. According to Pythagorean trigonometric identity

We substitute this into our equation

And we get a quadratic equation. We denote sin(x)=t

This gives sin (x) = 1/2 or sin (x) = 2. The latter is invalid, as the sine only gets values between -1 and 1. Then only sin (x) = 1/2 is valid, so the solutions are

where n is an integer.

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