

Solving Trigonometric Equations
Presentation
•
Mathematics
•
9th - 12th Grade
•
Hard
James Gonzalez
FREE Resource
5 Slides • 13 Questions
1
Solve for Principal Values:
2
Sum and Differences Identities and Solving Trig Equations
3
Multiple Choice
1) 2sin2x - sinx - 1 = 0
x = 3π/2, x = π/2
x = π/4, x = π/2
x = 5π/4, x= 3π/2
x = π/2, x = π/6
4
Multiple Choice
2) 3cosx - 2 = 0
Solve x over the interval [0,2π]
5
Multiple Choice
3)4tan2x - 1 = 0
Solve x over the interval [0,2π]
6
Multiple Choice
4) 2sin2x = 5sinx + 3
x = 7π/6, 5π/6
x = 7π/6, 11π/6
x = 5π/6, 11π/6
x = 5π/6, π/6
7
Multiple Choice
5) 2sin2x - 3sinx + 1 = 0
x = π/3, 4π/3
x = π/4, 3π/4
x = π/2, 7π/6
x = π/6, π/2
8
Find exact values:
9
Multiple Choice
Use the boundaries 0 < x < π/2, 0 < y < π/2
6) Find the exact value of sin (x+y) if sin x= 3/5 and sin y= 7/8
3√15/40 + 21/40
√34/40 + 3√113/40
10
Multiple Choice
7) Find cos 180
0
-1
1
11
Multiple Choice
8) tan (x+y) if csc x= 5/3 and cos y= 8/17
100/-13
85/96
72/85
-13/96
12
Double Identities:
13
Multiple Choice
cos2θ/cosθ - sinθ = cosθ + sinθ
9) What is the last step before the answer?
cos2θ/sinθ + sin2θ/cosθ
cosθ/cos2θ + sinθ/cos2θ
cos2θ/cosθ + sin2θ/sinθ
sinθ/cos2θ + cosθ/sin2θ
14
Multiple Choice
10) Find sin2θ, cos2θ, and tan2θ if cotθ = 13/7
180 < θ < 270
sin2θ = -13/-40, cos2θ = -218/169, tan2θ = 13/218
sin2θ = 147/218, cos2θ = -40/218, tan2θ = 147/-40
15
Verify (focus only on left side of equation):
16
Multiple Choice
sin (270 + A) = -cos A
11) What is the last step before verifying (=)?
-1(cos A) + 0(sin A)
1(cos A) + 0(sin A)
0(cos A) + 1(sinA)
0(cos A) + -1(sin A)
17
Multiple Select
tan (θ + 2π) = tan θ
12) What is the first and second step? (Choose 2 answers)
sin(θ+2π) /cos(θ+2π)
cos(θ+2π)
/sin(θ+2π)
cosθcos2π - sinθsin2π /sinθcos2π + cosθsin2π
sinθcos2π + cosθsin2π / cosθcos2π - sinθsin2π
18
Multiple Select
sec x - cos x = sinxtanx
13) What are one of the Identities used on the problem? (choose 2)
secx = 1/cosx
cotx = cosx/sinx
sin2x = 1 - cos2x
cos2x = 1 - sin2x
Solve for Principal Values:
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