8.5 Solving More Difficult Trigonometric Equations

Objective To use trigonometric identities or technology to solve more difficult trigonometric equations.

x

y

[Solution] Take the fourth root of both sides to obtain:

cos(2 )=±2

3

From the unit circle, the solutions for 2 are 2 = ± + kπ, k any integer.

π 6

Example 1: To find all solutions of cos4(2 ) = .9

16

1 π 6-π 6

x = -2

3 x =2

3

ππ

Answer: = ± + k , for k is any integer.12

2

Example 2: Find all solutions of the trigonometric equation: tan2 + tan = 0.

The solutions for tan = 0 are the values = kπ, for k is any integer.

Therefore, tan = 0 or tan = -1.

tan2 + tan = 0 Original equation

tan (tan +1) = 0 Factor.

The solutions for tan = -1 are = - + kπ, for k is any integer.

π 4

Some trigonometric equations can be transformed into equations that have quadratic form.

Example 3: The trigonometric equation

2 cos2 – 3 sin – 3 = 0 .

2 sin2 + 3 sin + 1 = 0 implies that

= -π/2 + 2kπ, from sin = -1

(2 sin + 1)(sin + 1) = 0.

Therefore, 2 sin + 1 = 0 or sin + 1 = 0.

Solutions: = - + 2kπ and = + 2kπ, from sin = -π

67π 6

1 2

It follows that sin = - or sin = -1.1 2

[Solution] Use the Pythagorean identity:

2 (1 – sin2 ) – 3 sin – 3 = 0 .

Steps for Solving:

• Isolate the Trigonometric function.• Then solve for the angle exactly if the ratio

represents known special triangle ratios.• Or solve for the angle approximately using the

appropriate inverse trigonometric function.

Complete the List of Solutions:

If you are not restricted to a specific interval and are asked to give a complete list of solutions (general solution), then remember that adding on any integer multiple of 2π represents a co-terminal angle with the equivalent trigonometric ratio.

8 sin = 3(1 sin2 ) Use the Pythagorean Identity.

[Solution] Rewrite the equation in terms of only one trigonometric function.

Example 4: Solve 8 sin = 3 cos2 with in the interval

[0, 2π].

3 sin2 + 8 sin 3 = 0. A “quadratic” equation with sin x as the variable

Therefore, 3 sin 1 = 0 or sin + 3 = 0

(3 sin 1)(sin + 3) = 0 Factor.

= arcsin( ) 0.3398 and = π arcsin( ) 2.8107.1 3

1 3

Solutions: sin = or sin = -31 3

Example 5: 5cos2 + cos – 3 = 0 for 0 ≤ ≤ π.

This is the range of the inverse cosine function.

The solutions are: = cos 1(0.6810249 ) = 0.8216349 and = cos 1(0.8810249) = 2.6488206

Therefore, cos = 0.6810249 or –0.8810249.

Use the calculator to find values of in 0 ≤ ≤ π.

[Solution] The equation is quadratic. Let u = cos and solve 5u2 + u 3 = 0.

u = -1 ± = 0.6810249 or -0.881024961 10

Example 6: Solve a trigonometric equation by factoring.

tan x sin x sin x

tan x sin x sin x 0

sin x tan x 1 0

tan 1 0 or sin 0x x

245

2 24 4

(2 1)4

x k

x k k

k

[Solution]

tan 1 or sin 0x x

x k

4x k

Example 7: Solve tan2x – 3tanx – 4 = 0 in 0 ≤ x ≤ 2.

[Solution] Let u = tan x, then

x2 – 3x – 4 = 0(x – 4)(x + 1) = 0x – 4 = 0 or x + 1 = 0x = 4 or x = –1

tan x = 4 or tan x = –1 x = arctan4 or x = 3/4 (x = 76° or x = 135°)

x = + arctan4 or x = 7/4 ( x = 256° or x = 315°)

Example 8: Solve. 3sinx + 4 = 1/sinx in 0 ≤ x ≤ 2 .

[Solution] Let x = sin x, then

3x + 4 = 1/x

x(3x + 4) = x(1/x) 3x2 + 4x = 1(3x – 1)(x + 1) = 03x – 1= 0 or x + 1 = 0x = 1/3 or x = –1sin x = 1/3 or sin x = –1

x = arcsin(1/3) or x = 3/2( x = 19° or x = 270°)x = – arcsin(1/3)( x = 161°)

2 2 Solve sin sin cos for 0 2 .x x x x Example 9 :

2 2sin sin cosx x x

2 2sin sin 1 sinx x x 22sin sin 1 0x x

2sin 1 sin 1 0x x

2sin 1 0 sin 1 0x x 1

sin sin 12

x x

7 11,

6 6 2x x

[Solution]

Solve sin tan 3sin . x x xExample 10 :

sin tan 3sin x x x

sin tan 3sin =0x x x

sin tan 3 0x x

sin 0 tan 3 0x x

0,x

tan 3x

arctan 3, arctan 3x

[Solution]

Particular solutions are

2x karctan 3x k General solutions are

2x k x k

Example 11: Solve 2 cos x + sec x = 0

[Solution]

2

12cos 0

cos1

2cos 1 0cos

xx

xx

2 1cos ,

2x

1cos

2x

is not a real number,

thus the equation has no solution.

10

cos x

Therefore,22cos 1 0x

10

cos x

or

However, So, 22cos 1x

Example 12: Solve cos x + 1 = sin x in [0, 2]

[Solution]

3, or

2 2x x

2cos 1 = 1 cos

Since 1 cos 1, 0 cos 1 2 2Thus sin 0 sin 1 cos

2 2cos 2cos 1 1 cosx x 22cos 2cos 0x x

2cos (cos 1) 0x x

2cos 0 or cos 1 0x x

cos 0 or cos 1x x Since we squared the original equation we have to check our answer. The only solutions are /2 and .

Like solving algebraic equation, once we squared the original trigonometric equation, it may generate some extraneous solution. We need to check the solutions.

Book Example 5. Solve 2sin cos 1 for 0 < 360 .

2sin cos 1 2 2 2sin 1 cos sin 1 cos

22 1 cos cos 1

2

222 1 cos cos 1 2 24 1 cos cos 2cos 1

2 24 4cos cos 2cos 1 20 5cos 2cos 3

5cos 3 cos 1 0

5cos 3 0 cos 1 0 3

cos cos 15

53.3 ,306.9 180

Since we squared the original equation we have to check our answer.

2sin53.1 cos53.1 1 1.6 1.6

2sin180 cos180 1 0 0

2sin306.9 cos306.9 1 1.6 1.6

the only solutions are 53.1 and 180

Book Example 5. Solve 2sin cos 1 for 0 < 360 .

2sin cos 1 Since 1 cos 1, 0 cos 1 2

Then 2sin cos 1 0

2

222 1 cos cos 1 2 24 1 cos cos 2cos 1

2 24 4cos cos 2cos 1 20 5cos 2cos 3

5cos 3 cos 1 0

5cos 3 0 cos 1 0 3

cos cos 15

53.3 ,306.9 180

Since we squared the original equation we have to check our answer.

the only solutions are 53.1 and 180

2Thus sin 0 sin 1 cos

2sin53.1 cos53.1 1 1.6 1.6

2sin180 cos180 1 0 0

2sin306.9 cos306.9 1 1.6 1.6

a)

Reference Angle:

Therefore:

b)

The equation cannotbe factored. Therefore,use the quadratic equation to find the roots:

Reference Angles:

sin 13

tan b b ac

a

2 42

tan( ) ( ) ( )( )

( )

1 1 4 3 12 3

2

4 1sin sin 3 1sin

0 3398.

0 3398 2 8018. .and

3 1 02tan tan

tan . tan . 0 43 0 76or

0 4061. 0 6499.

2 7355 5 877 0 6499 3 7915. , . , . , .

Using a Graphing Calculator to Solve Trigonometric Equations0 2

y = 3sin -1

5.2.13

3 tan2 tan 1 0

Using a Graphing Calculator to Solve Trigonometric Equations

Therefore, and

Remark

Solving trigonometric equations is a long lasting task through out the entire trigonometry. After we learned the sum and difference of angles, double angles, triple angles, we will be able solve some much more difficult trigonometric equations.

Assignment

P. 326 #1 – 22