Trigonometric Equations

In quadratic form, using identities or linear in sine and cosine

Solving a Trig Equation in Quadratic Form

Solve the equation: 2sin2 θ – 3 sin θ + 1 = 0, 0 ≤ θ ≤ 2 Let sin θ equal some variable sin θ = a Factor this equation (2a – 1) (a – 1) = 0 Therefore a = ½ a = 1

Solving a Trig Equation in Quadratic Form

Now substitute sin θ back in for a sin θ = ½ sin θ = 1 Now do the inverse sin to find what θ

equals θ = sin-1 (½) θ = sin-1 1 θ = /6 and 5/6 θ = /2

Solving a Trig Equation in Quadratic Form

Solve the equation: (tan θ – 1)(sec θ – 1) = 0 tan θ – 1 = 0 sec θ – 1 = 0 tan θ = 1 sec θ = 1 θ = tan-1 1 θ = sec-1 1 θ = /4 and 5/4 θ = 0

Solving a Trig Equation Using Identities

In order to solve trig equations, we want to have a single trig word in the equation. We can use trig identities to accomplish this goal.

Solve the equation 3 cos θ + 3 = 2 sin2 θ Use the pythagorean identities to

change sin2 θ to cos θ

Solving a Trig Equation Using Identities

sin2 = 1 – cos2 θ Substituting into the equation 3 cos θ + 3 = 2(1 – cos2 θ) To solve a quadratic equation it must

be equal to 0 2cos2 θ + 3 cos θ + 1 = 0 Let cos θ = b

Solving a Trig Equation using Identities

2b2 + 3b + 1 = 0 (2b + 1) (b + 1) = 0 (2b + 1) = 0 b + 1 = 0 b = -½ b = -1 cos θ = -½ cos θ = -1 θ = 2/3, 4/3 θ =

Solving a Trig Equation Using Identities

cos2 θ – sin2 θ + sin θ = 0 1 – sin2 θ – sin2 θ + sin θ = 0 -2sin2 θ + sin θ + 1 = 0 2 sin2 θ – sin θ – 1 = 0 Let c = sin θ 2c2 – c – 1 = 0 (2c + 1) (c – 1) = 0

Solving a Trig Equation Using Identities

(2c + 1) = 0 c – 1 = 0 c = -½ c = 1 sin θ = -½ sin θ = 1 θ = + θ = θ = 4/3,

Solving a Trig Equation Using Identities

Solve the equation sin (2θ) sin θ = cos θ Substitute in the formula for sin 2θ (2sin θ cos θ)sin θ=cos θ 2sin2 θ cos θ – cos θ = 0 cos θ(2sin2 – 1) = 0 cos θ = 0 2sin2 θ=1

Solving a Trig Equation Using Identities

cos θ = 0

θ = 0, θ = /4,

1 2sin

22

Solving a Trig Equation Using Identities

sin θ cos θ = -½ This looks very much like the sin

double angle formula. The only thing missing is the two in front of it.

So . . . multiply both sides by 2 2 sin θ cos θ = -1 sin 2θ = -1 2 θ = sin-1 -1

Solving a Trig Equation Using Identities

2θ = 3 θ = 3/4 2θ = 3

Solving a Trig Equation Linear in sin θ and cos θ

sin θ + cos θ = 1 There is nothing I can substitute in for

in this problem. The best way to solve this equation is to force a pythagorean identity by squaring both sides.

(sin θ + cos θ)2 = 12

Solving a Trig Equation Linear in sin θ and cos θ

sin2 θ + 2sin θ cos θ + cos2 θ = 1 2sin θ cos θ + 1 = 1 2sin θ cos θ = 0 sin 2θ = 0 2θ = 2θ = θ = θ = θ = θ = 3

Solving a Trig Equation Linear in sin θ and cos θ

Since we squared both sides, these answers may not all be correct (when you square a negative number it becomes positive).

In the original equation, there were no terms that were squared

Solving a Trig Equation Linear in sin θ and cos θ

Check: Does sin 0 + cos 0 = 1? Does sin cos Does sin cos Does sin 3cos 3

Solving a Trig Equation Linear in sin θ and cos θ

sec θ = tan θ + cot θ sec2 θ = (tan θ + cot θ)2

sec2 θ = tan2 θ + 2 tan θ cot θ + cot2 θ sec2 θ = tan2 θ + 2 + cot2 θ sec2 θ – tan2 θ = 2 + cot2 θ 1 = 2 + cot2 θ -1 = cot2 θ

Solving a Trig Equation Linear in sin θ and cos θ

is undefined (can’t take the square root of a negative number).

Solving Trig Equations Using a Graphing Utility

Solve 5 sin x + x = 3. Express the solution(s) rounded to two decimal places.

Put 5 sin x + x on y1

Put 3 on y2

Graph using the window 0 ≤ θ ≤2 Find the intersection point(s)

Word Problems

Page 519 problem 58