How to Solve Trigonometric Equations Using Identities

Some equations have multiple terms with trigonometric functions. In that case, you have to use the trigonometric identities to group the x’s. Here are some examples of using trigonometric identities in equations.

Example 1

Equations in the Form acos v + bsin v = 0

In this case, you divide by cos x0 on both sides of the equation to get an expression with tan x. This works because tan v = sin v cos v. Lastly, in this example, you have to check the case where cos 2x = 0 to make sure you don’t miss any solutions.

You are solving the equation

4cos2x + 4sin2x = 0
(1)

for x [0, 4π).

4 cos 2x + 4 sin 2x = 0|÷ 4 cos 2x + sin 2x = 0|÷ cos 2x0 cos 2x cos 2x + sin 2x cos 2x = 0 1 + tan 2x = 0 tan 2x = 1 2x = 3π 4 + n π x = 3π 8 + n π 2 In the interval [0, 4π), that gives you the solutions
x {3π 8 , 7π 8 , 11π 8 , 15π 8 , 19π 8 , 23π 8 , 27π 8 , 31π 8 }

x {3π 8 , 7π 8 , 11π 8 , 15π 8 , 19π 8 , 23π 8 , 27π 8 , 31π 8 } .

Now you have to check if cos 2x = 0 gives any solutions. Note that cos 2x = 0 is the same as sin 2x = 1 or sin 2x = 1:

sin 2x = 1 sin 2x = 1 2x = π 2 + n 2π 2x = 3π 2 + n 2π x = π 4 + n π x = 3π 4 + n π

sin 2x = 1 sin 2x = 1 2x = π 2 + n 2π 2x = 3π 2 + n 2π x = π 4 + n π x = 3π 4 + n π

From this you get the following solutions in the interval [0, 4π):

x {π 4, 3π 4 , 5π 4 , 7π 4 , 9π 4 , 11π 4 , 13π 4 , 15π 4 }

Now you have to test these answers in the main equation (1):

x = π 4:

LHS = 4 cos (2 π 4 ) + 4 sin (2 π 4 ) = 4 0 + 4 1 = 4 0 = RHS

x = 3π 4 :

LHS = 4 cos (2 3π 4 )) + 4 sin (2 3π 4 ) = 4 0 + 4 (1) = 4 0 = RHS

x = 5π 4 :

LHS = 4 cos (2 5π 4 ) + 4 sin (2 5π 4 ) = 4 0 + 4 1 = 4 0 = RHS

x = 7π 4 :

LHS = 4 cos (2 7π 4 ) + 4 sin (2 7π 4 ) = 4 0 + 4 (1) = 4 0 = RHS

x = 9π 4 :

LHS = 4 cos (2 9π 4 ) + 4 sin (2 9π 4 ) = 4 0 + 4 1 = 4 0 = RHS

x = 11π 4 :

LHS = 4 cos (2 11π 4 ) + 4 sin (2 11π 4 ) = 4 0 + 4 (1) = 4 0 = RHS

x = 13π 4 :

LHS = 4 cos (2 13π 4 ) + 4 sin (2 13π 4 ) = 4 0 + 4 1 = 4 0 = RHS

x = 15π 4 :

LHS = 4 cos (2 15π 4 ) + 4 sin (2 15π 4 ) = 4 0 + 4 (1) = 4 0 = RHS

Equation (1) doesn’t have any more solutions, and you can conclude that the solutions in the interval [0, 4π) are:

x {3π 8 , 7π 8 , 11π 8 , 15π 8 , 19π 8 , 23π 8 , 27π 8 , 31π 8 }

x {3π 8 , 7π 8 , 11π 8 , 15π 8 , 19π 8 , 23π 8 , 27π 8 , 31π 8 } .

Note! This is a very cumbersome way to check whether cos 2x = 0 yields multiple solutions, but it always works! Another way of checking cos 2x = 0 is to look at the main equation and using what you know about trigonometric identities to analyse it.

Example 2

Equations in the form

a cos(2x) + b cos x + c = 0

require you to use the identities

cos(2x) = cos 2x sin 2x

and

sin 2x + cos 2x = 1.

Solve the equation

cos(2x) + cos x + 1 = 0, = x [0, 2π)

cos(2x) + cos x + 1 = 0 cos 2x sin 2x + cos x + 1 = 0 cos 2x (1 cos 2x) + cos x + 1 = 0 2 cos 2x + cos x = 0 cos x(2 cos x + 1) = 0 You use the zero product property and look at the two equations cos x = 0 and 2 cos x + 1 = 0. The equation cos x = 0 has the solutions x1 = π 2 + n 2π x2 = π 2 + n 2π

You rewrite the equation 2 cos x + 1 = 0 as the basic equation cos x = 1 2. It has the solutions

x3 = 2π 3 + n 2π x4 = 2π 3 + n 2π

That means the solutions for the interval [0, 2π) are

x {π 2, 2π 3 , 4π 3 , 3π 2 }

Example 3

Equations in the form

a cos 2x b sin x = c

require you to use the identity

cos 2x + sin 2x = 1

and substitution.

Solve the equation

2 cos 2x sin x = 1, = x [0, 2π)

2 (1 sin 2x) sin x = 1 2 2 sin 2x sin x = 1 2 sin 2x sin x + 1 = 0 You make the substitution u = sin x in this equation, which gives you
2u2 u + 1 = 0.

Solve the quadratic equation:

u = 1 ±(1)2 4 (2) 1 4 = 1 ±9 4

u = 1 ±(1)2 4 (2) 1 4 = 1 ±9 4

The solutions to this are

u1 = 1u2 = 1 2

Now you substitute sin x back in for u:

sin x = 1 x1 = 3π 2 + n 2π x2 = π 3π 2 + n 2π = π 2 + n 2π sin x = 1 2 x3 = π 6 + n 2π x4 = π π 6 + n 2π = 5π 6 + n 2π

sin x = 1 sin x = 1 2 x1 = 3π 2 + n 2π x3 = π 6 + n 2π x2 = π 3π 2 + n 2π x4 = π π 6 + n 2π = π 2 + n 2π = 5π 6 + n 2π

The solutions in the interval [0, 2π) are:

x {π 6, 5π 6 , 3π 2 }

Note that π 2 is not included among the solutions. That’s because it’s outside the interval we’re looking at.

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