Pc Solve Trig Equations

Solving Trigonometric Equations

1 2

sin x = is a trigonometric equation. x = 6π is one of infinitely many solutions of y = sin x. y -19π 6 -3π

-11π 6

-7π 6

π 6

1

5π 6

π

-2π -π

13π 6

17π 6

2π 3π

25π 6 4π

1 x y=2

-1

All the solutions for x can be expressed in the form of a general solution. π

π

x = 6 + 2k π and x = 5 6 + 2k π (k = 0, ±1, ± 2, ± 3,

).

Find the general solution for the equation sec θ = 2. 1 1 From cos θ = , it follows that cos θ = . sec θ 2 1 All values of θ for which cos θ = 2

are solutions of the equation. Two solutions are θ = ± π . 3

All angles that are coterminal with ± π are also solutions and 3 can be expressed by adding integer multiples of 2π.

π

y

cos( 3 P

1 + 2kπ) = 2

1 x

1 2

Q 1 -π cos( 3 + 2kπ) =

The general solution can be written as θ = ± π + 2kπ . 3

2

Example: Solve tan x = 1. The graph of y = 1 intersects the graph of y = tan x infinitely y many times. - π – 2π 4

-π–π 4

π 4

π + π π + 2π 4

π + 3π

4

4

y=1 -π

π

2π

3π

x

y = tan(x) x = -3π x = -π 2 2

x = π x = 3π x = 5π 2 2 2

Points of intersection are at x = π and every multiple of π added or 4 π subtracted from 4 . General solution: x = π + kπ for k any integer. 4

Example: Solve the equation 3sin x + 2 = sin x for − π ≤ x ≤ π . 2

2

y

3sin x + 2 = sin x 3sin x − sin x + 2 = 0 2sin x + 2 = 0 sin x = −

2 2

Collect like terms.

1

-π 1 4

x

y=-

x = −4π is the only solution in the interval − 2π ≤ x ≤ 2π .

2 2

Example: To find all solutions of cos4(2x) = 9 . 16

Take the fourth root of both sides to obtain: cos(2x)= ± 3 y 2

From the unit circle, the solutions for 2θ are 2θ = ± 6π + kπ, k any integer.

π π

1

π 6

-π

x

6

x=- 3 2

π + k ( π ), for k any integer. Answer: θ = ± 12 2

x=

3 2

Find all solutions of the trigonometric equation: tan2 θ + tan θ = 0. tan2 θ + tan θ = 0 tan θ (tan θ +1) = 0

Original equation Factor.

Therefore, tan θ = 0 or tan θ = -1. The solutions for tan θ = 0 are the values θ = kπ, for k any integer. The solutions for tan θ = 1 are θ = - 4π + kπ, for k any integer.

The trigonometric equation 2 sin2 θ + 3 sin θ + 1 = 0 is quadratic in form. 2 sin2 θ + 3 sin θ + 1 = 0 implies that (2 sin θ + 1)(sin θ + 1) = 0. Therefore, 2 sin θ + 1 = 0 or sin θ + 1 = 0. It follows that sin θ = - 1 or sin θ = -1. 2

Solutions: θ = - π + 2kπ and θ = 7π + 2kπ, from sin θ = - 1 6 6 2 θ = -π + 2kπ, from sin θ = -1

Example: Solve 8 sin θ = 3 cos2 θ with θ in the interval [0, 2π]. Rewrite the equation in terms of only one trigonometric function. 8 sin θ = 3(1− sin2 θ )

Use the Pythagorean Identity.

3 sin2 θ + 8 sin θ − 3 = 0.

A “quadratic” equation with sin x as the variable (3 sin θ − 1)(sin θ + 3) = 0 Factor. Therefore, 3 sin θ − 1 = 0 or sin θ + 3 = 0 Solutions: sin θ = 1 or sin θ = -3 3 θ = sin−1( 1 ) = 0.3398 and θ = π − sin−1( 1) = 2.8107. 3 3 s

Solve: 5cos2 θ + cos θ – 3 = 0 for 0 ≤ θ ≤ π. The equation is quadratic. Let y = cos θ and solve 5y2 + y − 3 = 0. y = (-1 ± 61 ) = 0.6810249 or -0.8810249 10 Therefore, cos θ = 0.6810249 or –0.8810249. Use the calculator to find values of θ in 0 ≤ θ ≤ π. This is the range of the inverse cosine function. The solutions are: θ = cos −1(0.6810249 ) = 0.8216349 and θ = cos −1(−0.8810249) = 2.6488206

Example: Find the intersection points of the graphs of y y = sin θ and y = cos θ. The two solutions for θ between 0 and 2π are π and 5π . 4

4

5 -π 4

π + kπ 4

1 -π 4

π + kπ 4

1

The graphs of y = sin θ and y = cos θ intersect at points where sin θ = cos θ. This is true only for 45-45-90 triangles. The general solution is θ = 4π + kπ, for k any integer.

x