When x is between 0 and 2pi
a) sinx=0.7
When theta is between negative pi and pi
a) 6sinthetacostheta-3costheta+2sintheta-1=0
When theta is between 0 and 2pi
b)3 cos^2theta-13costheta=sin^2theta-9
thanks in advance if you can help me
When x is between 0 and 2pi
a) sinx=0.7
When theta is between negative pi and pi
a) 6sinthetacostheta-3costheta+2sintheta-1=0
When theta is between 0 and 2pi
b)3 cos^2theta-13costheta=sin^2theta-9
thanks in advance if you can help me
Hello, Confuzzled!
When 0 ≤ x ≤ 2π
. . a) . sin x .= .0.7
You need a calculator for this one.
. . x .= .Inverse sine of (0.7) .≈ .0.7754, 2.3662 radians.
When -π ≤ θ ≤ π
. . a) . 6sinθcosθ - 3cosθ + 2sinθ - 1 .= .0
Factor: . 3cosθ(2sinθ - 1) + (2sinθ - 1) .= .0
Factor: . (2sinθ - 1)(3cosθ + 1) .= .0
We have: . 2sinθ - 1 .= .0 . → . sinθ = 1/2 . → . θ .≈ .0.524, 2.618
. . . .and: . 3cosθ + 1 .= .0 . → . cosθ = -1/3 . → . θ .≈ .1.911, -1.911
When 0 ≤ θ ≤ 2π
. . b) . 3cos²θ - 13cosθ .= .sin²θ - 9
We have: . . . .3cos²θ - 13cosθ .= .1 - cos²θ - 9
. . .Then: . 4cos²θ - 13cosθ + 8 .= .0
This does not factor; we need the Quadratic Formula.
. . . . . . . . . . . . . . __
. . . . . . . . . .13 ± √41
. . cosθ . = . ----------- . ≈ . 2.4254, 0.8246
. . . . . . . . . . . . 8
We have: . cosθ .= .2.4254 . . . no solution!
. . . .and: . cosθ .= .0.8246 . → . θ .≈ .0.6013, 5.6819 radians
Thanks for your help. I’ve attempted to answer the following questions but it seems that I get many answers and I’m not sure if I’m doing it right. Could u help me with the following?:
For theta being between 0 and 360
a) sin^2theta=4cos^2theta
For theta being between pi and 2pi
b) 4cos^2theta=3
Hello again, Confuzzled!
a) sin²θ = 4cos²θ . . 0° < θ < 360°
Divide both sides by cos²θ:
. . . sin²θ
. . . ------- .= .4
. . . cos²θ
. . . tan²θ .= .4
. . . .tanθ .= .±2
. . . . . . θ .≈ .63.4°, 116.6°, 243.4°, 296.6°
b) 4cos²θ = 3 . . π < θ < 2π
. . . . . . . . . . . . . . . .3
We have: . cos²θ .= .---
. . . . . . . . . . . . . . . .4
. . . . . . . . . . . . . . . . . _
. . . . . . . . . . . . . . . . .√3
Then: . . . . cosθ .= . ± ---
. . . . . . . . . . . . . . . . . 2
Therefore: . θ .= .7π/6, 11π/6