need help with some trig questions?

Question

Can someone with a graphing calculator solve this sums for me using the graphing calc?

o˚≤Θ<360˚
1. solution to the nearest tenth of a degree: sinΘ -1 = 4 sinΘ

0 ≤ x < 2π
2. 2sin²x - 6 sin x - 1 = 0

0 ≤ x < 2π
3. 2 cos²x - 10cos x +3 = 0

I am looking for the radian solution for qns 2 and 3..

Answer 1

1.
sinΘ -1 = 4 sinΘ
<=>
-1 = 3sinΘ
<=>
sinΘ = -1/3
=>
Θ = arcsin(-1/3) = -19.5°

To get the two solutions in the interval [0°;360°] apply the fllowing relations:
sinΘ = sin(180° - Θ)
sinΘ = sin(Θ + 360°)
Hence
Θ₁ = 180° - (-19.5°) = 199.55°
Θ₂ = -19.5° + 360° = 340.5°

2.
2sin²x - 6 sin x - 1 = 0
<=>
sin²x - 3 sin x - 1/2 = 0
This is a quadratic equation of sin x with the solution:
sin x = 3/2 ± √((3/2)² + 1/2) = (3 ± √11) / 2
because -1 ≤ sin x ≤ 1
sin x = (3 - √11) / 2 = -0.158312...
<=>
x = arcsin(((3 - √11) / 2)) = -0.15898
The solutions in the interval [0;2π] are
x₁ = π - x = 3.3006
x₂ = x + 2π = 6.1242

3.
2 cos²x - 10 cos x + 3 = 0
<=>
cos²x - 5 cos x - 3/2 = 0
This is a quadratic equation of cos x with the solution:
cos x = 5/2 ± √((5/2)² - 3/2) = (5 ± √19) / 2
because -1 ≤ cos x ≤ 1
sin x = (5 - √19) / 2 = 0.32055
<=>
x = arccos(((5 - √19) / 2)) = 1.2445
Which is in the interval. For the second solution apply:
cos x = cos(2π - x)
Hence the solutions are:
x₁ = x = 1.2445
x₂ = 2π - x = 5.0387