Example 1 – Using a Trigonometric Identity Solve the equation 1 + sin = 2 cos2. Solution:

We first need to rewrite this equation so that it contains only one trigonometric function. To do this, we use a trigonometric identity:

1 + sin = 2 cos2

1 + sin = 2(1 – sin2 )

2 sin2 + sin – 1 = 0

(2 sin – 1) (sin + 1) = 0

Given equation

Pythagorean identity

Put all terms on one side

Factor

Example 1 – Solution

2 sin – 1 = 0 or sin + 1 = 0

sin = or sin = –1

= or =

Because sine has period 2, we get all the solutions of the equation by adding integer multiples of 2 to these solutions.

cont’d

Set each factor equal to 0

Solve for sin

Solve for sin in interval [0, 2)

Example 1 – Solution

Thus the solutions are

= + 2k = + 2k = + 2k

where k is any integer.

cont’d

Example 3 – Squaring and Using an Identity Solve the equation cos + 1 = sin in the interval

[0, 2).

Solution: To get an equation that involves either sine only or

cosine only, we square both sides and use a Pythagorean identity:

cos + 1 = sin

cos2 + 2 cos + 1 = sin2

cos2 + 2 cos + 1 = 1 – cos2

2 cos2 + 2 cos = 0

Given equation

Pythagorean identity

Square both sides

Simplify

Example 3 – Solution

2 cos = 0 or cos + 1 = 0

cos = 0 or cos = –1

= or =

Because we squared both sides, we need to check for extraneous solutions. From Check Your Answers we see that the solutions of the given equation are /2 and .

cont’d

Set each factor equal to 0

Solve for cos

Solve for in [0, 2)

Example 3 – Solution Check Your Answer:

= = =

cos + 1 = sin cos + 1 = sin cos + 1 = sin

0 + 1 = 1 0 + 1 ≟ –1 –1 + 1 = 0

cont’d

Equations with Trigonometric Functions of Multiples of Angles

Equations with Trigonometric Functions of Multiples of Angles

When solving trigonometric equations that involve functions of multiples of angles, we first solve for the multiple of the angle, then divide to solve for the angle.

Example 5 – A Trigonometric Equation Involving Multiple of an Angle Consider the equation 2 sin 3 – 1 = 0. (a) Find all solutions of the equation. (b) Find the solutions in the interval [0, 2). Solution: (a) We first isolate sin 3 and then solve for the

angle 3.

2 sin 3 – 1 = 0

2 sin 3 = 1

sin 3 =

Given equation

Add 1

Divide by 2

Example 5 – Solution

3 =

cont’d

Solve for 3 in the interval [0, 2) (see Figure 2)

Figure 2

Example 5 – Solution To get all solutions, we add integer multiples of 2

to these solutions. So the solutions are of the form

3 = + 2k 3 = + 2k

To solve for , we divide by 3 to get the solutions

where k is any integer.

cont’d

Example 5 – Solution

(b) The solutions from part (a) that are in the interval [0, 2) correspond to k = 0, 1, and 2. For all other values of k the corresponding values of lie outside this interval.

So the solutions in the interval [0, 2) are

cont’d

Example 6 – A Trigonometric Equation Involving a Half Angle

Consider the equation .

(a) Find all solutions of the equation.

(b) Find the solutions in the interval [0, 4).

Solution: (a) We start by isolating tan :

Given equation

Add 1

Example 6 – Solution

Since tangent has period , to get all solutions we add integer multiples of to this solution. So the solutions are of the form

Divide by

Solve for in the interval

cont’d

Example 6 – Solution

Multiplying by 2, we get the solutions

= + 2k

where k is any integer.

(b) The solutions from part (a) that are in the interval [0, 4) correspond to k = 0 and k = 1. For all other values of k the corresponding values of x lie outside this interval. Thus the solutions in the interval [0, 4) are

cont’d

Example 3A: Solving Trigonometric Equations with Trigonometric Identities

Use trigonometric identities to solve each equation.

tan2θ + sec2θ = 3 for 0 ≤ θ ≤ 2π.

tan2θ + (1 + tan2θ) – 3 = 0

2tan2θ – 2 = 0

tan2θ – 1 = 0

(tanθ – 1)(tanθ + 1) = 0

tanθ = 1 or tanθ = – 1

Substitute 1 + tan2θ for sec2θ by the Pythagorean identity.

Simplify. Divide by 2.

Factor.

Apply the Zero Product Property.

Example 3B: Solving Trigonometric Equations with Trigonometric Identities

Use trigonometric identities to solve each equation.

cos2θ = 1 + sin2θ for 0° ≤ θ ≤ 360°

(1 – sin2θ) – 1– sin2θ = 0

–2sin2θ = 0

sin2θ = 0

sinθ = 0

θ = 0° or 180° or 360°

Substitute 1 – sin2θ for cos2θ by the Pythagorean identity.

Simplify.

Divide both sides by – 2.

Take the square root of both sides.

Check It Out! Example 3a

Use trigonometric identities to solve each equation for the given domain.

4sin2θ + 4cosθ = 5

4(1 - cos2θ) + 4cosθ – 5 = 0

4cos2θ – 4cosθ + 1 = 0

(2cos2θ – 1)2 = 0

Substitute 1 – cos2θ for sin2θ by the Pythagorean identity.

Simplify.

Factor.

Take the square root of both sides and simplify.

Check It Out! Example 3b

Use trigonometric identities to solve each equation for the given domain.

sin2θ = – cosθ for 0 ≤ θ < 2

2cosθsinθ + cosθ = 0

cosθ(2sinθ + 1) = 0

Substitute 2cosθsinθ for sin2θ by the double-angle identity.

Factor.

Apply the Zero Product Property.

Example 4 – Finding Intersection Points

Find the values of x for which the graphs of f (x) = sin x and g (x) = cos x intersect.

Solution 1: GraphicalThe graphs intersect where f (x) = g (x). In Figure 1 we graph y1 = sin x and y2 = cos x on the same screen, for x between 0 and 2.

(a) (b)Figure 1

Example 4 – Solution 1

Using or the intersect command on the graphing calculator, we see that the two points of intersection in this interval occur where x 0.785 and x 3.927.

Since sine and cosine are periodic with period 2, the intersection points occur where

x 0.785 + 2k and x 3.927 + 2k

where k is any integer.

cont’d

Example 4 – Solution 2

Algebraic To find the exact solution, we set f (x) = g (x) and solve the resulting

equation algebraically:

sin x = cos x

Since the numbers x for which cos x = 0 are not solutions of the equation, we can divide both sides by cos x:

= 1

tan x = 1

cont’d

Equate functions

Divide by cos x

Reciprocal identity

Example 4 – Solution 2

The only solution of this equation in the interval (– /2, /2) is x = /4. Since tangent has period , we get all solutions of the equation by adding integer multiples of :

x = + k

where k is any integer. The graphs intersect for these values of x.

You should use your calculator to check that, rounded to three decimals, these are the same values that we obtained in Solution 1.

cont’d

Lesson Quiz

Solve cos2θ = 3sinθ + 2 for 0 ≤ 0 ≤ 2π.

HW 29page 396

#17,20, 24, 28, 32, 35-39odd, 79