Question

Solve the following for θ where θ is greater than or equal to 0 and less than 2π: cos^2θ=3cosθ

Answer 1

move everything to left side so it equals 0

factor out a cos

set each factor equal to 0 and solve

Answer 2

Would the answer for \[\theta \] be 0 then?

Answer 3

So if cosθ=0, it's just π/2 and 3π/2?

Answer 4

\(\bf cos(2 \theta) = 3cos(\theta) \implies cos(2 \theta)-3cos(\theta) = 0\\
\color{blue}{cos(2 \theta) = 2cos^2(\theta) -1}\\
cos(2 \theta)-3cos(\theta) = 0 \implies 2cos^2(\theta) -1-3cos(\theta) = 0\\
2cos^2(\theta)-3cos(\theta)-1=0 \implies 2x^2-3x-1=0\)

so solve it, I gather you may end up using the quadratic formula for that

Answer 5

yes its pi/2 and 3pi/2
im assuming its cos squared correct ?