Solve sin^2(x)=cos(x )for all solutions 0≤x<2π. Ask for X? Give your answers accurate to 2 decimal places, as a list separated by commas

1 Answer
Jun 20, 2018

51^@83, 308^@17

Explanation:

sin^2 x = cos x
(1 - cos^2 x) = cos x
cos^2 x + cos x - 1 = 0.
Solve this quadratic equation for cos x.
D = d^2 = b^2 - 4ac = 1 + 4 = 5 --> d = +- sqrt5
There are 2 real roots:
cos x = -b/(2a) +- d/(2a) = - 1/2 +- sqrt5/2 = (-1 +- sqrt5)/2
a. cos x = (-1 - sqrt5)/2 = -3.24/2 = -1.62 (rejected as < - 1)
b. cos x = (-1 + sqrt5)/2 = 0.62
Calculator and unit circle give 2 solutions for x -->
x = +- 51^@83
Note. x = - 51.83 is co-terminal to x = 360 - 51.83 = 308^@17
Answers for (0, 2pi)
51^@83, 308^@17
Check by calculator.
x = 308.17 --> sin^2 x = 0.62 --> cos x = 0.62. Proved