Solve each equation for 0<_ 0<_ 2pi(3.14)
A) 2cos^20 - sin0 -1=0
2cos^2 Ø - sinØ -1=0
2(1-sin^2 Ø) - sinØ - 1 = 0
-2sin^2 Ø - sinØ +1 = 0
2sin^2 Ø + sinØ - 1 = 0
(2sinØ + 1)(sinØ - 1) = 0
sinØ = -1/2 or sinØ = 1
case1: sinØ = -1/2
Ø must be in quads III or IV
Ø = 210 or Ø = 330°
or
Ø = 7π/6 or 11π/6
case2:
sinØ = 1
Ø = π/2
solve for theta:2cos(2theta-pi)= -2 ,for -2pi less or equal to theta less than 2pi
To solve the equation 2cos^2θ - sinθ - 1 = 0, we will first rewrite it as a quadratic equation in terms of cosθ.
Let's define a variable, x = cosθ. This means that x^2 = cos^2θ.
Now, substitute x^2 in the equation: 2x^2 - sinθ - 1 = 0.
Since sinθ = √(1 - cos^2θ), we can replace sinθ in the equation with √(1 - x^2): 2x^2 - √(1 - x^2) - 1 = 0.
To solve this quadratic equation, we can use the substitution method. Let u = 1 - x^2.
Now, substitute u in the equation: 2x^2 - √u - 1 = 0.
Rearrange the equation to isolate √u: √u = 2x^2 - 1.
Square both sides of the equation to remove the square root: u = (2x^2 - 1)^2.
Substitute back x^2 for cos^2θ: u = (2cos^2θ - 1)^2.
Remember that u = 1 - x^2, so replace it in the equation: 1 - x^2 = (2cos^2θ - 1)^2.
Expand and simplify the equation: 1 - x^2 = 4cos^4θ - 4cos^2θ + 1.
Rearrange the terms: 4cos^4θ - 4cos^2θ + x^2 = 0.
Finally, we have obtained a quadratic equation in terms of cosθ: 4cos^4θ - 4cos^2θ + x^2 = 0.
Now, you can solve this equation using various methods such as factoring, completing the square, or using the quadratic formula.