Question

Mohon bantuannyaa.
(No. 1) cos4x-5sin2x+2 < 0, 0≤x≤270
(No. 2) cos2x > cosx, 0≤x≤360
(No. 3) 1 < √3 sinx - cosx < √2, 0 (No. 4) √3 cosx - 3sinx + √3 > 0, 0

Answer 1

1. 1- 2(sin2x)^2 - sin2x = 0--->2(sin2x)^2 + sin2x - 1 = 0\
(sin2x +1)(2sin2x - 1) = 0
sin2x + 1 = 0--->sin2x = -1 ---> 2x = 270, 630 --->x = 135, 315
2sin2x - 1 = 0 ---> sin2x = 1/2 --->2x = 30, 150, 390, 510
---> x = 15, 75, 195, 255

2. 3 cos(2x) − cos x + 2 = 0
3 (2 cos²x − 1) − cos x + 2 = 0
6 cos²x − 3 − cos x + 2 = 0
6 cos²x − cos x − 1 = 0
(3 cosx + 1) (2 cosx − 1) = 0

3 cosx + 1 = 0
cosx = −1/3
x = arccos(−1/3), 360 − arccos(−1/3)
x = 109.47°, 250.53°

2 cosx − 1 = 0
cosx = 1/2
x = 60°, 300°

3. tidak tau :D

semoga membantu ^_^