Solving trigonometric equations: Give the two smallest solutions of cos(5theta)=0.1495, on [0,2pi). Where do I begin?

cos(5theta)=-0.1495cos(5θ)=−0.1495
given: [0,pi)[0,π)

I have no clue where to start here!!!

2 Answers
Sean
Jun 21, 2018

Please see below.

Explanation:

.

cos(5theta)=-0.1495cos(5θ)=−0.1495 Given [0,pi)[0,π)

arccos(-0.1495)=5thetaarccos(−0.1495)=5θ

5theta=1.72 Radians=98.6^@5θ=1.72Radians=98.6∘

theta=1.42/5=0.344 Radians=19.72^@θ=1.425=0.344Radians=19.72∘

If what is listed in small print is correct and your domain is [0, pi)[0,π) there is only one answer.

What you typed is 0.14950.1495 but in small print it is -0.1495−0.1495 and you typed [0,2pi)[0,2π) but in small print it is [0,pi)[0,π)

Nghi N.
Jun 21, 2018

2 smallest:
t = 16^@28; t = 55^@72t=16∘28;t=55∘72

Explanation:

cos 5t = 0.1495
Calculator and unit circle give 2 solutions for (5t):
5t = +- 81^@40 + k360^@5t=±81∘40+k360∘

a. 5t = 81^@40 + k360^@5t=81∘40+k360∘
t = 16^@28 + k72^@t=16∘28+k72∘
k = 0 --> t = 16.28; k = 1 --> t = 88.28; k = 2 --> t = 160.28;
k = 3 --> t = 232.28; k = 4 --> t = 304.28

b. 5t = - 81^@405t=−81∘40, or 5t = 278^@60 + k360^@5t=278∘60+k360∘ (co-terminal)
t = 55^@72 + k72^@t=55∘72+k72∘
k =0 --> t = 55.72; k = 1 --> t = 127.72; k = 2 --> t = 199.72;
k = 3 --> t = 271.72; k = 4 --> t = 343.72.
Therefor, for (0, 360), the 2 smallest values of t are:
t = 16^@28t=16∘28, and t = 55^@72t=55∘72

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