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
see your earlier post
Sure! Let's start with the first equation:
cos(2θ) = 4 - 3cos(θ)
To solve this equation, we can use a trigonometric identity: cos(2θ) = 2cos²(θ) - 1. Substituting this into the equation, we get:
2cos²(θ) - 1 = 4 - 3cos(θ)
Rearranging the terms, we have:
2cos²(θ) + 3cos(θ) - 5 = 0
Now, we have a quadratic equation in terms of cos(θ). To solve it, we can factor it or use the quadratic formula. In this case, factoring might be a bit complicated, so let's use the quadratic formula:
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
So, we have two possible solutions for cos(θ):
1) cos(θ) = (-3 + 7) / 4 = 1
2) cos(θ) = (-3 - 7) / 4 = -5/4
However, we need to find the values of θ. Since cos(θ) = 1 when θ = 0, and cos(θ) never takes the value of -5/4 in the interval 0 ≤ θ < 2π, we discard the second solution.
Therefore, the primary solution for cos(2θ) = 4 - 3cos(θ) is θ = 0.
Now, let's move on to the second equation:
cos(2θ)cos(θ) = sin(2θ)sin(θ)
Using trigonometric identities, we can rewrite this equation as:
cos(2θ)cos(θ) = 2sin(θ)cos(θ)cos(θ)
cos(2θ)cos(θ) - 2sin(θ)cos²(θ) = 0
Factor out cos(θ):
cos(θ)[cos(2θ) - 2sin(θ)cos(θ)] = 0
Setting each factor equal to zero, we have two equations to solve:
1) cos(θ) = 0
2) cos(2θ) - 2sin(θ)cos(θ) = 0
For the first equation, cos(θ) = 0 for θ = π/2 and θ = 3π/2.
For the second equation, we can use the identity cos(a+b) = cos(a)cos(b) - sin(a)sin(b):
cos(2θ) - 2sin(θ)cos(θ) = cos(θ + θ) - 2sin(θ)cos(θ)
= cos²(θ) - sin²(θ) - 2sin(θ)cos(θ)
= cos²(θ) - (1 - cos²(θ)) - 2sin(θ)cos(θ)
= 2cos²(θ) - 1 - 2sin(θ)cos(θ)
= 2cos²(θ) - 1 - sin(2θ)
Now, we have:
2cos²(θ) - 1 - sin(2θ) = 0
This is similar to the first equation we solved earlier. By rearranging the terms and applying the quadratic formula, we can find the solutions for this equation.
I hope this explanation helps! Let me know if you have any more questions.