find all values of degrees in the set (o,2 pi) satisfying cos of degree over 3 equal to square root of 3 over two
and -cosdegree + 2sindegres=-2
To find all values of degrees in the set (0, 2π) satisfying cos(θ/3) = √3/2, you can use the inverse cosine function, also known as arccosine.
1. Start by rearranging the equation cos(θ/3) = √3/2 to isolate θ/3:
θ/3 = arccos(√3/2)
2. Apply the inverse cosine function to both sides:
θ/3 = π/6 + 2πn or θ/3 = 11π/6 + 2πn (where n is an integer)
3. Multiply both sides by 3 to obtain θ:
θ = π/2 + 6πn or θ = 11π/2 + 6πn
Thus, the values of degrees in the set (0, 2π) satisfying cos(θ/3) = √3/2 are θ = π/2 + 6πn and θ = 11π/2 + 6πn, where n is an integer.
Now, let's solve the second equation: -cos(θ) + 2sin(θ) = -2.
1. Rearrange the equation to isolate either sine or cosine:
-cos(θ) + 2sin(θ) = -2
2sin(θ) = cos(θ) - 2
2. Square both sides to eliminate the sine term:
4sin^2(θ) = cos^2(θ) - 4cos(θ) + 4
3. Use the Pythagorean identity sin^2(θ) + cos^2(θ) = 1 to substitute for sin^2(θ):
4(1 - cos^2(θ)) = cos^2(θ) - 4cos(θ) + 4
4. Simplify and collect like terms:
4 - 4cos^2(θ) = 5cos^2(θ) - 4cos(θ)
5. Rearrange the equation to obtain a quadratic form:
9cos^2(θ) - 4cos(θ) - 4 = 0
This is a quadratic equation in terms of cos(θ). You can solve it using the quadratic formula or factoring.