Question

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

Answer 1

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.