How do you solve 8cosx+3=48cosx+3=4?

1 Answer
Nghi N
Dec 31, 2016

82^@82 + k(360^@)82∘82+k(360∘)
277^@18 + k(360^@)277∘18+k(360∘)

Explanation:

8cos x + 3 = 4
8cos x = 1 --> cos x = 1/8cosx=18
Use calculator and unit circle -->
cos x = 1/8cosx=18 --> x = +- 82^@82x=±82∘82
Note: the arc (-82.82) is co-terminal to arc (360 - 82.82) = 277^@18
Answers for (0, 360):
82^@82, 277^@1882∘82,277∘18
General answers:
82^@82 + 2k(180^@)82∘82+2k(180∘)
277^@18 + 2k(180^@)277∘18+2k(180∘)

Check by calculator.
cos (277.18) = 0.1249 = 1/8cos(277.18)=0.1249=18 .OK

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