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How to Solve Basic Trigonometric Equations

When solving trigonometric equations, it’s very important to include all the solutions! That means you have to be aware of the period of the function you are working with and how it influences the number of solutions.

Rule

Basic Trigonometric Equations

When a,b [1, 1] and n , the following is true:

sin x = a x = u + n 2π x = π u + n 2π cos x = b x = u + n 2π x = u + n 2π tan x = c x = u + n π

sin x = a x = u + n 2π x = π u + n 2π cos x = b x = u + n 2π x = u + n 2π tan x = c x = u + n π

Here, u represents the number you find for sin 1(a), cos 1(b) or tan 1(c).

Note! It’s very important to check which values x is allowed to have. It varies from problem to problem, and influences which values for n you can use in your answer.

Example 1

Solve the equation 4 cos (πx + 2π 3 ) = 2 for x (0, 2π)

You begin by transforming the equation to get the cos-term on its own: 4 cos (πx + 2π 3 ) = 2, cos (πx + 2π 3 ) = 1 2.

This has the solutions πx1 + 2π 3 = π 3 + n 2π, (1) πx2 + 2π 3 = π 3 + n 2π. (2)

First, you continue with (1): πx1 + 2π 3 = π 3 + n 2π πx1 = π 3 + n 2π x1 = 1 3 + 2n

Then you continue with (2): πx2 + 2π 3 = π 3 + n 2π πx2 = π + n 2π x2 = 1 + 2n

The problem tells you to find all the solutions that are in the interval x (0, 2π). You find these by considering x1 and x2 with respect to that interval.

Look at x1 = 1 3 + 2n first. If you insert n = 1, you get

x1 = 1 3 + 2 1 = 5 3,

which is in the interval. When you check n = 2, you get

x1 = 1 3 + 2 2 = 11 3 (0, 2π).

Then you check n = 3,

x1 = 1 3 + 2 3 = 17 3 ,

which is also in the interval. Now you notice that if you check n = 4, the answer will be outside the interval (0, 2π 6.28). That means you have found all the solutions for x1.

Now you have to do the same for x2 = 1 + 2n. The values still have to be in the interval for them to be a part of the solution. That gives you n = 1 x2 = 1 + 2 1 = 1 (0, 2π) n = 2 x2 = 1 + 2 2 = 3 (0, 2π) n = 3 x2 = 1 + 2 3 = 5 (0, 2π) n = 4 x2 = 1 + 2 4 = 7 (0, 2π)

As you can see, the last value is outside the interval. That means the solutions in the interval (0, 2π) are:

x {1, 5 3, 3, 11 3 , 5, 17 3 } .

Example 2

Solve the equation sin (2x π 3 ) = 1 2 for x [0, 2π)

The basic equation

sin (2x π 3 ) = 1 2

has the solutions 2x1 π 3 = sin 1 (1 2) + n 2π, (3) 2x2 π 3 = π sin 1 (1 2) + n 2π. (4)

First, you continue working on (3): 2x1 π 3 = π 6 + n 2π 2x1 = π 2 + n 2π| : 2 x1 = π 4 + n π

Then you work on (4): 2x2 π 3 = π π 6 + n 2π 2x2 = 7π 6 + n 2π| : 2 x = 7π 12 + n π

Now you have to find the solutions from x1 and x2. The values have to be in the interval x [0, 2π) for them to be one of the solutions. For x1 = π 4 + 2π, you have n = 0 x1 = π 4 + 0 = π 4, n = 1 x1 = π 4 + 1π = 5π 4 , n = 2 x1 = π 4 + 2π = 9π 4 ,

where 9π 4 is outside the interval. When you check x2 = 7π 12 + nπ, you get n = 0 x2 = 7π 12 + 0 = 7π 12, n = 1 x2 = 7π 12 + 1π = 19π 12 , n = 2 x2 = 7π 12 + 2π = 31π 12 ,

where 31π 12 is outside the interval. That means solutions in the interval [0, 2π) are:

x {π 4, 7π 12, 5π 4 , 19π 12 } .

Example 3

Solve the equation 3 tan (3x + 5π 6 ) = 3 for x

You solve the trigonometric equation for x:

23 tan (3x + 5π 6 ) = 3| : 3 tan (3x + 5π 6 ) = 1 3x + 5π 6 = tan 1(1) 3x + 5π 6 = π 4 + nπ 3x = π 4 5π 6 + nπ 3x = 3π 10π 12 + nπ| : 3 x = 7π 36 + nπ 3

3 tan (3x + 5π 6 ) = 3 | : 3 tan (3x + 5π 6 ) = 1 3x + 5π 6 = tan 1(1) 3x + 5π 6 = π 4 + nπ 3x = π 4 5π 6 + nπ 3x = 3π 10π 12 + nπ | : 3 x = 7π 36 + nπ 3

The solution to the equation is x = 7π 36 + nπ 3 for every n . Because x can be any real number in this example (x ), there is no specific interval you need to check n-values for. That means the solution is general.

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