multipy both sides by sintheta
1=sinTheta+cosTheta
This is only possible at Theta=0, 90, 360
Solve the equation in the interval [0•,360•]
Csc theta = 1+cot theta
3 answers
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.
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.
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)
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)
1 answer