5.3 Solving Trigonometric Equations

5.3 Solving Trigonometric Equations

Goal: To isolate the trigonometric function involved in the equation.

A). Solving a Trigonometric Equation

Example 1:
4 cos²(2x) - 2 = 0
4 cos²(2x) = 2
cos²(2x) =2/4
cos²(2x) = ½
cos (2x) = ± the square root of ½= ± (the square root of 2)/2
So now Let y = 2x
Now when does cos y = ±(the square root of 2)/2
y = 45°, 135°, 225°, 315°, π/4, 3π/4, 5π/4, 7π/4

so 2x = 45°
x = 22.5° or π/8

2x = 135°
x = 67.5° or 3π/8

2x = 225°
x = 112.5° or 5π/8

2x = 315°
x = 157.5° or 7π/8

and as you can see, this only takes you from 0 radians to π radians. If the solution is between [0, π), we would be done but usually it is between [0,2π) so we have to add the following solutions:

x = {9π/8, 11π/8, 13π/8, 15π/8}

Example 2:

(the square root of 2) (sin x )+ 1 = 0
(the square root of 2) sin x = -1
sin x = -1/ (the square root of 2) = -(the square root of 2)/2
sine is negative in the III and IV quadrant so

x = 225°, 315°, 5π/4, 7π/4

Example 3:

tan²x – 1 = 0
tan²x = 1
tan x = ±1

x = 45°, 135°, 225°, 315°, π/4, 3π/4, 5π/4, 7π/4

Example 4:

2 sin²x = 2 + cos x
2(1 - cos²x) = 2 + cos x
2 - 2cos²x = 2 + cos x
0 = 2cos²x + cos x
0 = cos x(2cos x + 1)
0 = cos x and 0 = 2cos x + 1

0 = cos x
x = 90°, 270°, π/2, 3π/2

2 cos x + 1 = 0
cos x = - ½
x = 120°, 240°, 2π/3, 4π/3

Check your solutions to check for erroneous solutions
either by algebra or using the graphing utility

B) Approximate the solution

Example 5:
csc²x = 3 csc x + 4
csc²x - 3 csc x - 4 = 0
(csc x - 4)(csc x + 1) = 0
csc x = 4 and csc x = -1

csc x = 4
(1/ sin x) = 4
sin x = 1/4
x ≈ 14.48° or .2527 radians

csc x = -1
x = 270° or 3π/2

Example 6:

4 sin x = cos x - 2
Use your graphing utility and place each side of the equation into 2 different equation functions.
f(x) = 4 sin x and g(x) = cos x - 2
You will find the intersection of the two equations are
(-2.390169347, -2.730717947)
(-.2614659803,-1.033987935)
(3.89301596, -2.730717947)
(6.021719327, -1.033987935)

or

(-136.9466°, -2.730717947)
(-14.9809 °, -1.033987935)
(223.05338 °, -2.730717947)
(345.0191 °, -1.033987935)

so therefore the solutions are:
x = {-136.9466 °, -14.9809 °, 223.05338 °, 345.0191 °}

x = {-2.390169347, -.2614659803, 3.89301596, 6.021719327}

Example 7:

cos x + sin x tan x = 2

cos x + (sin x)(sin x)/ (cos x) = 2

(cos²x + sin²x)/(cos x) = 2

1/(cos x) = 2

cos x = ½

x = {60 °, 300 °, π/3, 5π/3}

Example 8:

sec x csc x = 2 csc x
sec x csc x - 2 csc x = 0
csc x (sec x - 2 ) = 0

csc x = 0
1/ sin x = 0
sin x = undefined
no solution

sec x - 2 = 0
sec x = 2
1/ cos x =2/1
cos x = ½
x = {60 °, 300 °, π/3, 5π/3}

5.3 homework #45 - page 400; 3 - 59 (by 4's), 73, 76