**General Solutions of Trigonometric Equations**

sin θ = 0 cos θ = 0 |
θ = nπ, n ϵ lθ = (2n + 1) π/2, n ϵ l |

tan θ = 0 sin θ = sinα |
θ = nπ, n ϵ l θ = nπ + (-1)ⁿ, α ϵ [-π/2, π/2], n ϵ l |

cos θ = cosα tan θ = tan α |
θ = 2nπ ± α, α ϵ [0, π], n ϵ l θ = nπ + α, α ϵ (-π/2, π/2), n ϵ l |

sin² θ = sin²α cos² θ = cos²α tan² θ = tan²α |
θ = nπ ± α, n ϵ l |

sin θ = 1 |
θ = (4n + 1) π/2, n ϵ l |

cos θ = 1 |
θ = 2nπ, n ϵ l |

cos θ = -1 |
θ = (2n + 1) π, n ϵ l |

sinx = sinα and cosx = cosα |
x = 2nπ + α |

1) If θ is an odd multiple of π/2 i.e, when θ = (2n + 1) π/2, n ϵ l then secθ and tanθ are not defined

2) If θ is an even multiple of π/2 i.e, when θ = nπ, n ϵ l then cosec θ and cot θ are not defined

**Example:** Solve the equation tan θ + sec θ = √3

**Solution: **tan θ + sec θ = √3 … (1)

Then (sin θ)/ (cos θ) +1/ (cos θ) = √3 … (2) (∵ sin θ/cos θ = tan θ and 1/cos θ = secθ)

\(\frac{\sin \theta +1}{\cos \theta }=\sqrt{3}\)

Sinθ + 1 = √3 cosθ

½ Sinθ – √3/2 cosθ = -1/2

Sin(π/6) sinθ – cos(π/6) cosθ = cos(π/3)

We can use the cos (A + B) formula

cos (θ + π/6) = cos π/3

General solution θ + π/6 = 2nπ ± π/3, n ∈ I

Taking positive sign, θ + π/6 = 2nπ + π/3

⇒ θ = 2nπ + π/6

Taking negative sign, θ + π/6 = 2nπ – π/3

⇒ θ = 2nπ – π/2

i.e. θ = (4n – 1) π/2.

But the solution obtained is correct only if, cos θ ≠ 0 otherwise (2) is not defined.

i.e. θ ≠ odd multiple of π/2

⇒ θ ≠ (4n – 1) π/2.

Hence the general solution will be θ = 2nπ + π/6 only where n = 0, ±1, ±2, …