How do you find all solutions of the equation in the interval [0,2pi) given cos^2x-3sinx=-1?

1 Answer
Nghi N.
Mar 6, 2017

34^@06; 145^@94

Explanation:

Substitute in the equation cos^2 x by (1 - sin^2 x, and bring the equation to standard form.
1 - sin^2 x - 3sin x = - 1
- sin^2 x - 3sin x + 2 = 0
Solve this quadratic equation for sin x.
D = d^2 = b^2 - 4ac = 9 + 8 = 17 --> d = +- sqrt17
There are 2 real roots:
sin x = -b/(2a) +- d/(2a) = - 3/2 +- sqrt17/2
sin x = (-3 + sqrt17)/2 = (- 3 + 4.12)/2 = 0.56, and
sin x = (-3 - sqrt17)/2 = - 7.12/2 = - 3.56 (rejected as < - 1)

sin x = 0.56 --> calculator gives --> x = 34^@06, and the
unit circle gives another x = 180 - 34.06 = 145^@94
Answers for (0, 360):
34^@06; 145^@94
Check:
x = 145.94 --> cos x = - 0.83 --> cos^2 x = 0.69 --> 3sin x = 1.68 -->
cos^2 x - 3sin x = 0.69 - 1.68 = - 1. Proved.

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