Question

what is formula to find general solution in trignometry

Answer 1

Trigonometrical equation

General Solution

sin θ = 0 

 Then θ = nπ

cos θ = 0

 θ = (nπ + π/2)

tan θ = 0

 θ = nπ

sin θ = 1

 θ = (2nπ + π/2) = (4n+1)π/2

cos θ = 1

 θ = 2nπ

sin θ = sin α

 θ = nπ + (-1)nα, where α ∈ [-π/2, π/2]

cos θ = cos α

 θ = 2nπ ± α, where α ∈ (0, π]

tan θ = tan α

 θ = nπ + α, where α ∈ (-π/2 , π/2]

sin2 θ = sin2 α

 θ = nπ ± α

cos2 θ = cos2 α

 θ = nπ ± α

tan2 θ = tan2 α

 θ = nπ ± α

Answer 2

1. If sin θ = 0 then θ = nπ, where n = zero or any integer.

2. If sin θ = sin ∝ then θ = nπ + (-1)^n ∝, where n = zero or any integer.

3. If sin θ = 1 then θ = (4n + 1)π/2, where n = zero or any integer.

4. If sin θ = -1 then θ = (4n - 1)π/2, where n = zero or any integer.

5. If cos θ = 0 then θ = (2n + 1)π/2, where n = zero or any integer.

6. If cos θ = cos ∝ then θ = 2nπ ± ∝, where n = zero or any integer.

7. If cos θ = 1 then θ = 2nπ, where n = zero or any integer.

8. If cos θ = -1 then θ = (2n + 1)π, where n = zero or any integer.

9. If tan θ = 0 then θ = nπ, where n = zero or any integer.

10. If tan θ = tan ∝ than θ = nπ + ∝, where n = zero or any integer.

11. If sin^2 θ = sin2 ∝ then θ = nπ ± ∝, where n = zero or any integer.

12. If cos^2 θ = cos2 ∝ then θ = nπ ± ∝, where n = zero or any integer.

13. If tan^2 θ = tan2 ∝ than θ = nπ ± ∝, where n = zero or any integer.

14. If a cos θ + b sin θ = c then θ  = 2nπ + ∝ ±  β, where cos β = c/$\frac{c} \sqrt{a^{2}+ b^{2}  }$, cos ∝ = $\frac{a}{\sqrt{a^{2} + b^{2} } }$ and sin ∝ = , where n = zero or any integer.