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.