**Principal Solution and General Solution of Trigonometric Functions :**

Here we are going to see how to find principal solution and general solution of trigonometric functions.

**Principal Solution**

The smallest numerical value of unknown angle satisfying the equation in the interval [0, 2π] (or) [−π, π] is called a principal solution.

Principal value of sine function lies in the interval

[−π/2, π/2]

and hence lies in I quadrant or IV quadrant.

Principal value of cosine function is in

[0, π]

and hence in I quadrant or II quadrant.

Principal value of tangent function is in

(-π/2, π/2)

and hence in I quadrant or IV quadrant.

**General Solution :**

The solution of a trigonometric equation giving all the admissible values obtained with the help of periodicity of a trigonometric function is called the general solution of the equation.

Trigonometric equation sin θ = 0 cos θ = 0 tan θ = 0 sin θ = sinα, where α ∈ [−π/2, π/2] cos θ = cos α, where α ∈ [0,π] tan θ = tanα, where α ∈ (−π/2, π/2) |
General solution θ = nπ; n ∈ Z θ = (2n + 1) π/2; n ∈ Z θ = nπ; n ∈ Z θ = nπ + (−1) θ = 2nπ ± α, n ∈ Z θ = nπ + α, n ∈ Z |

**Question 1 :**

Find the principal solution and general solutions of the following:

(i) sin θ = −1/√2

**Solution :**

** sin θ = −1/√2**

θ = sin ^{-1}(−1/√2)

**Principal solution :**

We have to choose the principal solution between the interval [−π/2, π/2] .

θ = (-π/4)

**General solution :**

**θ = nπ + (−1) ^{n} α, n ∈ Z**

** θ = nπ + (−1)^{n} **(-π/4)

** (ii) **cot θ = √3

**Solution :**

cot θ = √3

1/tan θ = 1/√3

Taking reciprocals on both sides

tan θ = √3

**Principal solution :**

We have to choose the principal solution between the interval (−π/2, π/2)

θ = π/6

**General solution :**

**θ = nπ + α, n ∈ Z**

** θ = nπ + (**π/6)

**(iii) tan** θ = -1/√3

**Solution :**

tan θ = -1/√3

**Principal solution :**

We have to choose the principal solution between the interval (−π/2, π/2)

θ = -π/6

**General solution :**

**θ = nπ + α, n ∈ Z**

** θ = nπ + (-**π/6)

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