How can I solve this? Thank you in advance!!

Question

For 0 ≤ θ < 2π, find the solution(s) to the equation: -3cos(θ) = 2sin² (θ) - 3

Answer 1

This is a tricky one!

I had to go back through my precalc notes to find how to solve the problem.

There is a nifty strategy in solving trig equations that look like this that involves converting the equation into a quadratic and solving for the variable. First thing you will want to do is to convert that sin²θ into a 1-cos²θ. After that, you want to get everything on the same side of the equation equal to zero. This leaves you with the equation -3cosθ-2(1-cos²θ)+3 = 0.

This is where we use the strategy I mentioned. Sub any variable (I'm going to use x) you would like in for cosθ and solve as you would for any other quadratic equation. The equation then turns into (after multiplying through the two, taking care of the arithmetic, and subbing in my x) 2x²-3x+1=0. We can factor that quadratic to (2x-1)(x-1)=0. We have our zeros at 1/2 and 1. Remember, our variable represents cosine, so your solution is all the points on the interval between 0 and 2π where cosine is 1/2 and 1.

Good luck!

Answer 2

-3 cos θ = 2 sin² θ - 3

-3 cos θ = 2 (1 - cos² θ) - 3

Set equal to zero and use Algebra to solve for θ.

Answer 3

sin²θ = 1 - cos²θ

So, we have 2(1 - cos²θ) + 3cosθ - 3 = 0

2cos²θ - 3cosθ + 1 = 0

(2cosθ - 1)(cosθ - 1) = 0

cosθ = 1/2 or cosθ = 1

cosθ = 1 when θ = 0

cosθ = 1/2 when θ = π/3 or 5π/3

Solution set: {0, π/3, 5π/3}

Answer 4

We want to use a pythagorean ID (the fundamental trig ID) on sin²(Θ) in order to convert the equation to a quadratic in cosine:

-3cos(Θ) = 2(1 - cos²(Θ)) - 3

2cos²Θ - 3cosΘ + 1 = 0

(2cosΘ - 1)(cosΘ - 1) = 0

2cosΘ - 1 = 0 or cosΘ - 1 = 0

cosΘ = 1/2 or cosΘ = 1. Θ = π/3 or 5π/3 or 0