Using the t-formula to solve this equation [check my working out] (1 Viewer)

[Leffife]

Solve: √3 sinθ + cosθ = 1 for 0≤θ≤2π

My working out:

√3 [(2t)/(1+t²)] + [(1-t²)/(1+t²)] = 1

[(2√3.t + 1 - t²)/(1+t²)] = 1

2√3.t + 1 - t² = 1 + t²
2t² - 2√3.t = 0
2t(t - √3) = 0

t = 0 or t = √3

∴ tan(θ/2) = 0 or tan(θ/2) = √3
∴ θ/2 = 0 , π/3 , 4π/3
∴ θ = 0, 2π/3

The answer says 0, 2π/3 , 2π.

I'm not sure how they ended up with 2π, but solving it. I just need to see how they ended up getting there.

Thank you.

 

[barbernator]

all trig functions are periodic, so their values repeat after 2pi. therefore the value for 0 is equal to the value for 2pi so they are both solutions within the given domain.
also, tan(pi)=tan(0)=0, you just forgot one solution for that

 

[Carrotsticks]

all trig functions are periodic, so their values repeat after 2pi. therefore the value for 0 is equal to the value for 2pi so they are both solutions within the given domain

This.

Also there's the standard checking for 180 degrees etc, but really not necessary in this case.

 

[Leffife]

Yup gotcha, thanks. I should pay more attention.