Find all primary solutions (i.e. 0 ≤ θ < 2π ) of the equation cos(2θ ) = 4 − 3 cos(θ ).
Find all primary solutions (i.e. 0 ≤ θ < 2π ) of the equation cos(2θ )cos(θ ) = sin(2θ )sin(θ ).
Please can somone help and show all work Thank you these two examples would help me immensely
remember your double-angle formulas:
cos(2θ) = 4 − 3 cos(θ)
2cos^2(θ)-1 = 4 - 3cosθ
2cos^2(θ)+3cosθ - 5 = 0
(2cosθ-1)(cosθ+5) = 0
cosθ = -5 has no solution
cosθ = 1/2 has solutions at
θ = π/3 and 5π/3
and other basic trig identities:
cos(2θ)cosθ = sin(2θ)sinθ
(2cos^2(θ)-1)cosθ = 2sinθcosθsinθ
(2cos^2(θ)-1)cosθ = 2sin^2(θ)cosθ
(2cos^2(θ)-1)cosθ = 2(1-cos^2(θ))cosθ
2cos^3θ - cosθ = 2cosθ-2cos^3θ
4cos^3θ-3cosθ = 0
cosθ(4cos^2θ-3) = 0
cosθ = 0: θ = π/2 or 3π/2
cosθ = ±√3/2: θ = π/6, 5π/6, 7π/6, 11π/6
Or for the 2nd
Cos 2x cosx -sin 2xsinx = 0
Cos(2x + x) = 0
cos 3x = 0
3x = pi/2 or 3x = 3pi/2
x = pi/6 or x = pi/2
The period of cos 3x = 2pi/3
So add multiples of that together Steve's other answers
cos(x+pi)-sin(x-pi)=0 on the interval[0,2pi]
Sure! I can help you with both examples. Let's start with the first one:
To find all primary solutions of the equation cos(2θ) = 4 − 3 cos(θ), we can use trigonometric identities to simplify the equation. In this case, we can use the double angle formula for cosine:
cos(2θ) = 2cos²(θ) - 1
So, we can rewrite the equation as:
2cos²(θ) - 1 = 4 - 3cos(θ)
Rearranging the equation, we get:
2cos²(θ) + 3cos(θ) - 5 = 0
Now, this equation is in quadratic form. We can solve it by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:
The quadratic formula states that for an equation of the form ax² + bx + c = 0, the solutions are given by:
x = (-b ± √(b² - 4ac)) / (2a)
In our case, a = 2, b = 3, and c = -5. Substituting these values into the quadratic formula, we get:
cos(θ) = (-3 ± √(3² - 4(2)(-5))) / (2(2))
cos(θ) = (-3 ± √(9 + 40)) / 4
cos(θ) = (-3 ± √49) / 4
cos(θ) = (-3 ± 7) / 4
Simplifying further, we have two possibilities:
1) cos(θ) = (-3 + 7) / 4
cos(θ) = 4 / 4
cos(θ) = 1
2) cos(θ) = (-3 - 7) / 4
cos(θ) = -10 / 4
cos(θ) = -5/2
Now, to find the values of θ that satisfy these solutions, we can use the inverse cosine function (also known as arccos or cos⁻¹).
1) cos(θ) = 1
θ = cos⁻¹(1)
θ = 0
2) cos(θ) = -5/2
This is not possible since the range of the cosine function is -1 to 1.
Therefore, the only primary solution is θ = 0.
For the second example, let's solve the equation cos(2θ)cos(θ) = sin(2θ)sin(θ):
First, let's simplify the equation using trigonometric identities. We can use the double angle formula for cosine and sine:
cos(2θ) = 2cos²(θ) - 1
sin(2θ) = 2sin(θ)cos(θ)
Substituting these identities into our equation, we get:
(2cos²(θ) - 1)cos(θ) = (2sin(θ)cos(θ))sin(θ)
Expanding and rearranging, we have:
2cos³(θ) - cos(θ) = 2sin²(θ)cos(θ)
Now, let's use the Pythagorean identity sin²(θ) + cos²(θ) = 1 to substitute sin²(θ) with 1 - cos²(θ):
2cos³(θ) - cos(θ) = 2(1 - cos²(θ))cos(θ)
2cos³(θ) - cos(θ) = 2cos(θ) - 2cos³(θ)
Rearranging the terms, we get:
4cos³(θ) - 3cos(θ) = 0
To solve this equation, we can factor out the common factor of cos(θ):
cos(θ)(4cos²(θ) - 3) = 0
This equation is satisfied when cos(θ) = 0 or when 4cos²(θ) - 3 = 0.
1) cos(θ) = 0
θ = π/2 or 3π/2
2) 4cos²(θ) - 3 = 0
4cos²(θ) = 3
cos²(θ) = 3/4
Taking the square root of both sides, we have:
cos(θ) = ±√(3/4)
cos(θ) = ±√3/2
The values of θ that satisfy this equation are:
θ = arccos(√3/2) or θ = arccos(-√3/2)
Using the inverse cosine function, we find:
θ = π/6 or 11π/6
Therefore, the primary solutions for this equation are θ = π/2, 3π/2, π/6, and 11π/6.
I hope this explanation helps! Let me know if you have any more questions.