Posted by **Stacey grahaM** on Tuesday, May 4, 2010 at 9:35am.

Solve the equation in the interval [0•,360•]

Csc theta = 1+cot theta

- Trig -
**bobpursley**, Tuesday, May 4, 2010 at 10:03am
multipy both sides by sintheta

1=sinTheta+cosTheta

This is only possible at Theta=0, 90, 360

- Trig -
**MathMate**, Tuesday, May 4, 2010 at 10:11am
Reduce to sin and cosines:

1/sinθ=(sinθ+cosθ)/sinθ

Thus, if θ ≠ 0,π 2π ...

we get

sinθ+cosθ=1

Taking advantage of symmetry about π/4, where sinπ/4=cosπ/4, substitute θ=φ-π/4:

sin(φ-π/4)+cos(φ-π/4)=1

Expanding by sum/difference formulae,

sinφcosπ/4-cosφsinπ/4 + cosφcosπ/4+sinφsinπ/4=1

Since sinπ/4=cosπ/4, we cancel terms in cosφ to get

2sinπ/4 sinφ=1

φ=arcsin(sqrt(2)/2)=±π/4

θ=0 or π/2

The first value has been rejected since the beginning, so θ=π/2.

- Trig - correction -
**MathMate**, Tuesday, May 4, 2010 at 10:23am
from

φ=arcsin(sqrt(2)/2)=π/4 or 3π/4 ± 2kπ

θ=φ-π/4=0 or π/2 ±2kπ

Since θ=0 has been rejected since the beginning, we are left with

θ=π/2 (for solution between 0 and 360)

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