Solve cos 2 ( x ) = 8 sin ( x ) for all x in the interval 0 ≤ x < 2 π : x =

Answer 1

Use the double-angle identity for cosine that includes sin². This will allow us to write the equation as a quadratic in sinx which we can then solve:

1 - 2sin²x = 8sinx

2sin²x + 8sinx - 1 = 0 Here the discriminant, b² - 4ac = 72 which is not a perfect square so not factorable.

Using quadratic formula: sinx = (-8 ± √72)/4 = -2 ± 3√2/2

One of those solutions is NOT between -1 and 1 so won't yield a solution. However, -2 + 3√2/2 ~ .1213

sin^-1(.1213) ~ .1216 and there is a 2nd solution in QII : π - .1216 ~ 3.02

x ~ .1216 or 3.0200