Question

Question #6bcdd

Answer 1

`x = +- pi/3 + 2kpi`

Explanation:

Solve the trig equation:
`3cos x = 2sin^2 x`
Replace `sin^2 x` by `(1 - cos^2 x)` -->
`3cos x = 2 - 2cos^2 x`
`2cos^2 x + 3cos x - 2 = 0`
Solve this quadratic equation in cos x by using the improved quadratic formula (Google, Socratic Search)
`D = d^2 = b^2 - 4ac = 9 + 16 = 25` --> `d = +- 5`
There are 2 real roots:
`cos x = - b/(2a) +- d/(2a) = -3/4 +- 5/4` -->
`cos x = -8/4 = -2` (rejected as < - 1)
`cos x = 2/4 = 1/2`
Trig table and unit circle give 2 solutions:
`x = +- pi/3 + 2kpi`
Check.
`x = pi/3` --> `cos x = 1/2` --> `3cos x = 3/2` -->
`sin x = sqrt3/2` --> `sin^2 x = 3/4` --> `2sin^2 x = 3/2`. Proved.