Question

How do you solve `5cos2x+sinx=4` in the interval [0,360]?

Answer 1

`21^@72; 158^@28; 195^@66; 344^@44`

Explanation:

Use the trig identity: `cos 2x = 1 - 2sin^2 x`.
`5(1 - 2sin^2 x) + sin x - 4 = 0`
`5 - 10sin^2 x + sin x - 4 = 0`
`- 10sin^2 x + sin x + 1 = 0`
Solve this quadratic equation in sin x by using the improved quadratic formula (Socratic Search)
`D = d^2 = b^2 - 4ac = 1 + 40 = 41` --> `d = +- sqrt41 = +- 6.40`
There are 2 real roots
`sin x = -b/(2a) +- d/(2a) = 1/20 +- 6.40/20`
sin x = 7.40/20 = 0.37 and sin x = - 5.40/20 = - 0.27
Use calculator and trig unit circle -->
a. sin x = 0.37 --> arc `x = 21^@72` and arc
`x = 180 - 21.72 = 158^@28`
b. sin x = - 0.27 --> arc `x = -15.66, or 344^@44` (co-terminal)
and arc `x = 180 + 15.66 = 195^@66`
Answers for `(0, 360)`
`21^@72, 158^@28, 195^@66, 344^@44`