Question

I just dont understand these.
2 cos^2 θ − 5 cos θ + 2 = 0
sin^2 θ = 2 sin θ + 3
cos 2θ = 5 sin θ − 2
PLEASE HELP!!!

Answer 1

Solve the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to three decimal places where appropriate. If there is no solution, enter NO SOLUTION.)

Answer 2

I will show you the first, you can try the other two.

2 cos^2 θ − 5 cos θ + 2 = 0

This is basically a quadratic equation in cos(theta). Let t = cos(theta)

2t^2 - 5t + 2 = 0
2t^2 - 4t - t + 2 = 0
2t(t - 2) - 1(t - 2) = 0
(2t - 1)(t - 2) = 0
t = 1/2 or t = 2
Put t back as cos(theta)
cos(theta) = 1/2 or cos(theta) = 2. The second can't be true because cos is always less than or equal to 1 and so you can discard that solution
cos(theta) = 1/2 or theta = 60 degree or pi/3 radians.
Since cos is cyclic, the function repeats itself every multiple of 2(pi).
So theta = pi/3 +/- k(2(pi))

Answer 3

ok thank you

Answer 4

you are welcome.

Answer 5

Q= theta.
ok so for the next one would it be sin^2Q-2sinQ+3?

Answer 6

sin^2 θ = 2 sin θ + 3

Let t = sin(theta)

t^2 = 2t + 3
t^2 - 2t - 3 = 0. Solve for t. And then put back t as sin(theta) and then solve for theta.

For problem 3)
Use the double angle formula for cosines:
cos(2A) = cos²A − sin²A = 2 cos²A − 1 = 1 − 2 sin²A
The one that will be useful here is: cos(2A) = 1 − 2 sin²A
Then follow the same procedure of temporarily replacing sin(theta) with t, solving for t and then putting back sin(theta) in the place of t.

When you solve for t always be mindful that sine, cosine cannot be more than 1 or less than -1. So any solutions of t that is not in that range should be discarded.

Also they ask you to use k as the integer. So any answer you get for theta you remember to add plus/minus k2(pi) where k = 0, 1, 2, 3, etc.

Answer 7

Oh i was trying to do it step by step to make sure i was doing it right. Now i see how its -3