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Hi, can you please help me with these trig questions?

0 votes

1) 3 csc^2θ=4 find all solutions of the equation 2)a) 2 sin 3θ +1=0 b) find the solutions in the interval [0, 2pi) 3)a) sinθ = -0.6 Solve the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to two decimal places where appropriate.) (the answer should be in radians) b)list six specific solutions thank you!

asked Nov 26, 2014 in TRIGONOMETRY by anonymous

5 Answers

+1 vote

1)

3csc²θ=4

csc²θ=4/3

cscθ=√(4/3)

cscθ = 2/√3

 Substitute : cscθ = 1/sinθ

1/sinθ = 2/√3

sinθ = (√3)/2

sinθ = (√3)/2

sinθ = sin(π/3)

General solution for sin(θ) = sin(α) is θ = kπ + (-1)kα

θ = kπ + (-1)k(π/3)   where k is integer.

Solution : θ = kπ + (-1)k(π/3).

answered Nov 26, 2014 by Shalom Scholar
edited Nov 26, 2014 by steve
+1 vote

2a)

2 sin3θ + 1 = 0

Given interval is [0 , 2π)

2 sin3θ = -1

sin3θ = -½

sin3θ = sin(-π/6)

  General solution for sin(β) = sin(α) is β = kπ + (-1)kα

  Here β = 3θ , α = -π/6

3θ = kπ + (-1)k(-π/6)

θ = [kπ + (-1)k(-π/6)] /3

θ = kπ/3 + (-1)k(-π/(6*3))

θ = (π/3)k + (-1)k(-π/18)

Solution : θ = (π/3)k + (-1)k(-π/18).

answered Nov 26, 2014 by Shalom Scholar
edited Nov 26, 2014 by Shalom
+1 vote

2b)

θ = (π/3)k + (-1)k(-π/18)   ( From Solution2a )

For k = 0 ⇒ θ = (π/3)0 + (-1)0(-π/18) = -π/18 is out of given range [0,2π).

For k = 1 ⇒ θ = (π/3)1 + (-1)1(-π/18)

                     = (π/3)+(-1)(-π/18) = (π/3)+(π/18) = 7π/18

For k = 2 ⇒ θ = (π/3)2 + (-1)2(-π/18)

                     = (π/3)2+(1)(-π/18) = (2π/3)-(π/18) = 11π/18

For k = 3 ⇒ θ = (π/3)3 + (-1)3(-π/18)

                     = (π/3)3+(-1)(-π/18) = (3π/3)+(π/18) = 19π/18

For k = 4 ⇒ θ = (π/3)4 + (-1)4(-π/18)

                     = (π/3)4+(1)(-π/18) = (4π/3)-(π/18) = 23π/18

For k = 5 ⇒ θ = (π/3)5 + (-1)5(-π/18)

                     = (π/3)5+(-1)(-π/18) = (5π/3)+(π/18) = 31π/18

For k = 6 ⇒ θ = (π/3)6 + (-1)6(-π/18)

                     = (π/3)6+(1)(-π/18) = (6π/3)-(π/18) = 35π/18

For k = 7 ⇒ θ = (π/3)7 + (-1)7(-π/18)

                     = (π/3)7+(-1)(-π/18) = (7π/3)+(π/18) = 43π/18 is out of given range [0,2π).

So, for k > 7, x will be out of range [0,2π).

Solution :

x = { 7π/18 , 11π/18 , 19π/18 , 23π/18 , 31π/18 , 35π/18 }.

answered Nov 26, 2014 by Shalom Scholar
edited Nov 26, 2014 by Shalom
+1 vote

3a)

sinθ = -0.6

sinθ = -0.6

θ = - 36.8699°

θ = - 0.6435011

θ = - 0.64

General solution : θ = kπ + (-1)k(- 0.64)  ,where k is integer.

Solution : θ = kπ + (-1)k(- 0.64).

answered Nov 26, 2014 by Shalom Scholar
edited Nov 26, 2014 by Shalom
+1 vote

3a)

θ = kπ + (-1)k(- 0.64).   ( From solution 3a )

 

For k = 0 ⇒ 0*π + (-1)0(- 0.64) = 0 -  0.64 = - 0.64

For k = 1 ⇒ θ = 1π + (-1)1(- 0.64)

                       = π + (-1)(- 0.64) = 3.14 + 0.64 = 3.78

For k = 2 ⇒ θ = 2π + (-1)2(- 0.64)

                       = 2(3.14) + (1)(- 0.64) = 6.28 - 0.64 = 5.64

For k = 3 ⇒ θ = 3π + (-1)3(- 0.64)

                       = 3(3.14)  + (-1)(- 0.64) = 9.42 + 0.64 = 10.06

For k = 4 ⇒ θ = 4π + (-1)4(- 0.64)

                       = 4(3.14) + (1)(- 0.64) = 12.56 - 0.64 = 11.92

For k = 5 ⇒ θ = 5π + (-1)5(- 0.64)

                       = 5(3.14)  + (-1)(- 0.64) = 15.7 + 0.64 = 16.34

Solution :

x = { -0.64 , 3.78 , 5.64 , 10.06 , 11.92 , 16.34 }.

answered Nov 26, 2014 by Shalom Scholar

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