Question

MEDALS! solve the equation (answers in between 0 and 360 degrees). 1) sin^2 x-1=cos2 x and 2) 2cos^2 x+3sinx-3=0

Answer 1

answers are 1) 30, 150 degrees and 2) 135, 315 degrees but how do you do it?

Answer 2

can u please draw the equation in first case!

Answer 3

i'm supposed to do this without drawing.

Answer 4

\[\sin ^{2}x-1=\cos2x\]
\[\sin ^{2}x-1= 2\cos ^{2}x-1 \implies \tan ^{2}x=2\]
\[\implies x = \tan^{-1} 2, -\tan^{-1} 2 \]

Answer 5

i posted the answers, but i don't get what u did. did u get the answer?

Answer 6

1) sin^2 x - 1 = cos 2x. Replace cos 2x by (1 - 2sin^2 x)
sin^2 x - 1 = 1 - 2sin^2 x
3sin^2 x = 2 -> sin^2 x = 2/3 -> sin x = V2/V3 = V6/3 or -V6/3
Calculator gives x = .....
2) 2cos^2 x + 3sin x - 3 = 0. Replave cos^2 x by (1 - sin^2 x)
2 - 2sin^2 x + 3sin x - 3 = 0
- 2sin^2 x + 3sin x -1 = 0. This is a quadratic equation. Since a + b + c = 0, one real root is sin x = 1 and the other is sin x = -1/2.
a) sin x = 1 = sin Pi/2 -> x = .....
b) sin x = -1/2 . Trig Table or calculator gives: -1/2 = sin 7Pi/6 and 11Pi/6
sin x = sin (7Pi/6) --> x = ....
sin x = sin (11Pi/6) --> x = ....