Aim: Trig Equations with Multiple Functions Course: Alg. 2 & Trig.

Aim: How do we solve trig equations involving more than one trig function?

Do Now:

Find all values of in the interval 0 < < 360o that satisfy the equation

2 sin2 + sin = 1.

Aim: Trig Equations with Multiple Functions Course: Alg. 2 & Trig.

Solving Multiple Trig Function Equations

Solve for in the interval 0º ≤ ≤ 360º:2cos2 – sin = 1

cos2 = 1 – sin2

alternate form of Pythagorean ID

substitute: 2(1 – sin2 ) – sin = 1

2 – 2sin2 – sin = 1

-2sin2 – sin + 1 = 0

2sin2 + sin – 1 = 0

(2sin – 1)(sin + 1) = 0

(2sin – 1) = 0 (sin + 1) = 0

sin = 1/2 sin = -1 = 30º or 150º = 270º{30º,150º, 270º}

standard form/factor/solve:

Aim: Trig Equations with Multiple Functions Course: Alg. 2 & Trig.

Solving Multiple Trig Function Equations

Solve for to the nearest degree in the interval 0º ≤ ≤ 360º:2sec2 – 3tan - 5 = 0

sec2 = 1 + tan2

alternate form of Pythagorean ID

substitute: 2(1 + tan2 ) – 3tan - 5 = 0

2 + 2tan2 – 3tan - 5 = 0

2tan2 – 3tan - 3 = 0

standard form/factor/solve:

x b b2 4ac

2aa = 2, b = -3, c = -3

2( 3) ( 3) 4(2)( 3)

2(2)x

3 33

4x

tan = 2.19, tan = -0.69

= 65o, 145o, 245o, 3250

Aim: Trig Equations with Multiple Functions Course: Alg. 2 & Trig.

Model Problems

2

tan 2cot 1

7cos 1 6sec

sec 4 2tan

Find to the nearest degree, all values of in the interval 00 < < 3600 that satisfy the given equation.

Aim: Trig Equations with Multiple Functions Course: Alg. 2 & Trig.

Model Problems

2

2sin 3csc 5

2sin 1 csc

13tan 2

cot

Find to the nearest degree, all values of in the interval 00 < < 3600 that satisfy the given equation.

Aim: Trig Equations with Multiple Functions Course: Alg. 2 & Trig.

Aim: Trig Equations with Multiple Functions Course: Alg. 2 & Trig.