Yahoo Answers: Answers and Comments for 1. Solve the equation 5.5(sin^2 x) = 16cos x + 10.5 on the interval 0<x<7? [Mathematics]
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From Anonymous
enUS
Mon, 26 Nov 2018 17:25:40 +0000
3
Yahoo Answers: Answers and Comments for 1. Solve the equation 5.5(sin^2 x) = 16cos x + 10.5 on the interval 0<x<7? [Mathematics]
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From rotchm: From sin² = 1  cos² you have,
5.5( 1  c²) ...
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https://answers.yahoo.com/question/index?qid=20181126172540AAshXgP
Mon, 26 Nov 2018 17:33:36 +0000
From sin² = 1  cos² you have,
5.5( 1  c²) = 16c + 10.5, where c = cos(x).
This is a quadratic in c, so just solve for c.
That is, multiply each side by 2. Then put all terms on the same side.
Show the answer here....
Having found the value(s) for cos(x), just do arccos each side. Don forget that arccos has two values...
Done!
Show your steps here if need be and we can further guide you.
I hope no one will spoil you the answer thereby depriving you from your personal enhancement; that would be very inconsiderate of them.

From Mathmom:
Rewrite in terms of cos x only:
16 cos x +...
https://answers.yahoo.com/question/index?qid=20181126172540AAshXgP
https://answers.yahoo.com/question/index?qid=20181126172540AAshXgP
Mon, 26 Nov 2018 18:20:27 +0000
Rewrite in terms of cos x only:
16 cos x + 10.5 = 5.5(sin²x)
16 cos x + 10.5 = 5.5(1−cos²x)
16 cos x + 10.5 = 5.5 − 5.5 cos²x
5.5 cos²x + 16 cos x + 5 = 0
Now we can use quadratic formula to solve for cos x
cos x = (−16 ± √(16²−4(5.5)(5))) / (2(5.5))
cos x = (−16 ± √146) / 11
cos x = −2.5530042, −0.3560867
Since −1 ≤ cos x ≤ 1 for all real x, then
cos x = −0.3560867
x = 2nπ ± cos⁻¹(−0.3560867), where n is some integer
Find all values of x between 0 and 7