sin tan cos

Suppose the range of x : 0 ≤ x ≤ 360

1. cos 2x = 0.4
2x = 66.4, 294, 426 or 654
x = 33.2, 147, 213 or 327

2. √3 tan(x + 30) = 2
tan (x + 30) = 2/√3
x + 30 = 49.1 or 229
x = 19.1 or 199

3. √2 cos^2 x - cos x = 0
cos x(√2 cos x - 1) = 0
cos x = 0 or cos x = 1/√2
x = 90, 270 or x = 45, 315
The solution for x are 45, 90, 270 or 315

4. 2 sin^2 x - 5 sin x - 3 =0
(2 sin x + 1)(sin x - 3) = 0
sin x = -1/2 or sin x = 3 (rej.)
x = 210 or 330

5. √3 tan x = 2 sin x
√3 sin x = 2 sin x cos x
√3 sin x - 2 sin x cos x = 0
sin x(√3 - 2 cos x) = 0
sin x = 0 or √3 = 2 cos x
x = 0,180 or cos x = √3 / 2
x = 0,180 or x = 30, 330
The solution for x are 0, 30, 180 or 330

6. L.H.S. = tan (270 + θ)
= tan [90 + (180 + θ)]
Let a = 180 + θ
we have tan (90 + a)
= -1/tan a
= -1/tan (180 + θ)
= -1/tan θ

1. cos 2x = 0.4
2x = 66.4
x=33.2
2. √3tan(x+30) = 2
tan(x+30) = 2/√3
x+30 = 49.1
x = 19.1
3.√2cos^2 x - cosx = 0
cosx(√2cosx-1) = 0
cosx = 0 or cosx = 1/√2
x = 90 or 45
5.√3tanx = 2sinx
√3/cosx = 2
cosx = √3/2
x = 30
6.tan(270+θ) = -1/tanθ
tan(270+θ) = -tan(90-θ)
-tanθ = -tanθ