Solving Equations?

Question

1) can u solve the equation 2 cos 2θ = 4 - 5 cos θ

for values of θ between 0 degrees and 360 degrees.

note* the symbol θ is the greek symbol theta
could pls show step by step...

2) solve the equation:

5 * 4^(2x+1) - 10 * 4^x + 1 = 0

Note* could you please show step by step..

Answer 1

2) Note that 4^(2x+1) = 4^(2x)*4^1 = 4*4^(2x)
5*4^(2x+1) = 5*4*4^(2x) = 20*4^(2x)

5 * 4^(2x+1) - 10 * 4^x + 1 = 0
20*4^(2x) - 10*4^x + 1 = 0
Let y = 4^x, then y^2 = 4^(2x)
20y^2 - 10y + 1 = 0
Now use the quadratic formula to solve for y.
You will get two answers.
Set each one equal to 4^x and solve for x.

Answer 2

1) can u solve the equation 2 cos 2θ = 4 - 5 cos θ
for values of θ between 0 degrees and 360 degrees.

Use cos 2θ = 2cos^2θ -1
So, 2(2cos^2θ - 1) = 4 - 5 cosθ
4cos^2θ -2 = 4 - 5cosθ
4cos^2θ +5cosθ -6 = 0
(4cosθ -3)(cosθ+2)= 0
4cosθ -3 =0
cosθ =3/4
cosθ+2=0
cosθ = -2
θ = arccos(3/4) = 41.41 degrees or 318.59 degrees
θ = arccos (-2) = not possible/

2) solve the equation:

5 * 4^(2x+1) - 10 * 4^x + 1 = 0

Answer 3

Try going to: purplemath.com