Essential Question: How do we find the non-calculator solution to inverse sin and cosine functions?

8-3: Algebraic Solutions to Trig EquationsSolving Basic Cosine Equations

Example 1: Solve cos x = 0.6 Using the graphing calculator gets us the first solution cos-1 (0.6) = 0.9273 To find the second solution, we need to

use the identity rule of cosine cos(-x) = cos(x) This tells us that our second solution is at

x = -0.9273 Remember, cosine is a cyclical wave, so all

solutions are given by: x = 0.9273 + 2k and

x = -0.9273 + 2k

x

1

2

–1

–2

y

2

2

8-3: Algebraic Solutions to Trig EquationsSolving Basic Sine Equations

Example 2: Solve sin x = -0.75 Using the graphing calculator gets us the first solution sin-1 (-0.75) = -0.8481 To find the second solution, we need to

use the identity rule of sine sin( - x) = sin(x) This tells us that our second solution is at

x = 3.1416 – (-0.8481) = 3.9897 Remember, sine is a cyclical wave, so all

solutions are given by: x = -0.8481 + 2k and

x = 3.9897 + 2k (because the cycle is 2, 3.9897 -2.2935)

2

2

x

1

2

–1

–2

y

8-3: Algebraic Solutions to Trig EquationsSolving Basic Tangent Equations

Example 3: Solve tan x = 3 Using the graphing calculator gets us the solution tan-1 (3) = 1.2490 There is no second solution on a tangent

function, but remember, tangent is a cyclical wave, so all solutions are given by: x = 1.2490 + k

2

2

x

1

2

3

4

5

–1

–2

–3

y

8-3: Algebraic Solutions to Trig EquationsUsing the Solution Algorithm

Example 4: Solve 8 cos x – 1 = 0 Isolate the trigonometric equation

8 cos x = 1 cos x = 1/8

Use the inverse cosine to find the first solution cos-1 (1/8) = x x = 1.4455

Use the identity rule of cosine [cos(-x) = cos(x)] for the other x value x = -1.4455

All solutions are given by:x = 1.4455 + 2k and x = -1.4455 + 2k

8-3: Algebraic Solutions to Trig EquationsSolving Basic Equations with Special Values

Example 5: Solve sin u = exactly, without using a calculator. Find the first value by either:a) Use the table of values (restricted sin/cos/tan functions) you should

have copied – and started to memorize – by now.b) Use the calculator in degree mode, and convert degrees to radiansc) Use the calculator in radian mode, and divide your answer by

x = /4

Use the identity rule of sine [sin( - x) = sin(x)] for the other x value x = - /4 = 3/4

All solutions are given by:x = /4 + 2k and x = 3/4 + 2k

2

2

8-3: Algebraic Solutions to Trig EquationsAssignment

Page 545Problems 1 – 21, odd problems

Essential Question: How do we find the non-calculator solution to inverse sin and cosine functions?

8-3: Algebraic Solutions to Trig EquationsSolving basic equations with substitution

Example 6: Solve sin 2x = exactly, without using a calculator. Let u = 2x, this gets us into a basic equation

(we actually solved it yesterday)

sin u = u = /4 + 2k and u = 3/4 + 2k

Substitute 2x back in for u, then solve for x 2x = /4 + 2k and 2x = 3/4 + 2k x = /8 + k and x = 3/8 + k

2

2

2

2

8-3: Algebraic Solutions to Trig EquationsFactoring Trigonometric Equations

Example 7: Solve 3 sin2 x – sin x – 2 = 0 in the interval [-, ] Let u = sin x. This gets us into a quadratic equation

3u2 – u – 2 = 0 Factor

(3u + 2)(u – 1) = 0 Set each parenthesis = 0 and solve for u

u = - 2/3 and u = 1 Substitute sin x back in for u

sin x = - 2/3 and sin x = 1

Continued next slide

8-3: Algebraic Solutions to Trig EquationsFactoring Trigonometric Equations (Continued)

Example 7: Solve 3 sin2 x – sin x – 2 = 0 in the interval [-, ] sin x = - 2/3 and sin x = 1

Use the inverse sin function to solve for the 1st answer in each equation x = -.7297 + 2k and x = /2 + 2k

Use the identity rule of sine to find the alternate values x = 3.1416 – (-.7297) = 3.8713 x = – /2 = /2

3.8713 is outside the defined interval ([-, ]), but if you subtract a revolution (3.8713 – 6.2832), -2.4119 is within the interval.

This gives us our final solutions x = -2.4119 + 2k x = -.7297 + 2k x = /2 + 2k

8-3: Algebraic Solutions to Trig EquationsFactoring Trigonometric Equations #2

Example 8: Solve tan x cos2 x = tan x Write the expression = 0

tan x cos2 x – tan x = 0 Take out the GCF

tan x (cos2 x – 1) = 0 The right-part of the equation can be factored

tan x (cos x – 1)(cos x + 1) = 0 Set each parenthesis = 0 and solve

tan x = 0 cos x = 1 cos x = -1 Use the inverse trigonometric functions to solve

x = 0 + k x = 0 + 2k x = + 2k

8-3: Algebraic Solutions to Trig EquationsFactoring Trigonometric Equations #2

Solutions from the previous slide: x = 0 + k x = 0 + 2k x = + 2k

So where are those solutions?

Meaning all the solutions can be summed up as: x = 0 + k

0 0

2 3

2 4 5

3 6 7

8-3: Algebraic Solutions to Trig EquationsIdentities and Factoring

Example 9: -10 cos2 x – 3 sin x + 9 = 0 Use the Pythagorean identity to write everything in terms of sin

sin2 x + cos2 x = 1 cos2 x = 1 – sin2 x

Replace the cos2 x -10(1 – sin2 x) – 3 sin x + 9 = 0

Distribute 10 sin2 x – 10 – 3 sin x + 9 = 0 (combine like terms) 10 sin2 x – 3 sin x – 1 = 0

Factor (5 sin x + 1)(2 sin x – 1) = 0

Continued next slide

8-3: Algebraic Solutions to Trig EquationsIdentities and Factoring (Continued)

Example 9: -10 cos2 x – 3 sin x + 9 = 0(5 sin x + 1)(2 sin x – 1) = 0

Set each parenthesis equal to 0 and solve sin x = – 1/5 sin x = 1/2

Use the inverse sine function to solve for x (1st values) x = -0.2014 x = /6

Use the identity rule of sine to find the alternate values x = 3.1416 – (-0.2014) = 3.3430 x = – /6 = 5/6

Our final solutions: x = -0.2014 + 2k x = /6 + 2k

x = 3.1416 + 2k x = 5/6 + 2k

8-3: Algebraic Solutions to Trig EquationsAssignment

Page 546Problems 23 – 53, odd problems