# HOW TO SOLVE TRIGONOMETRIC EQUATIONS

Solving trigonometrical equations is nothing but finding all replacements for the variable θ that make the equations true.

A solution of a trigonometrical equation is the value of the unknown angle that satisfies the equation. A trigonometrical equation may have infinite number of solutions.

Principal Solution :

The solution in which the absolute value of the angle is the least is called principal solution.

General Solution :

The expression involving integer ‘n’ which gives all solutions of a  trigonometrical equation is called the general solution.   Example 1 :

Solve : 2cos2θ + 3sinθ = 0.

Solution :

2cos2θ + 3sinθ = 0

First we have to convert the given function in one trigonometric ratio.

cos2θ = 1 - sin2θ

2(1 - sin2θ) + 3sinθ = 0

2 - 2sin2θ + 3sinθ = 0

Multiply the whole equation by (-)

2sin2θ - 3sinθ - 2 = 0

2sin2θ - 4sinθ + sinθ - 2 = 0

2sinθ(sinθ - 2) + 1(sinθ - 2) = 0

(2sinθ + 1) (sinθ - 2) = 0

 2sinθ + 1 = 0Subtract 1 on both sides2sinθ + 1 - 1 = 0 - 12sinθ = -1Divide by 2 on both sidessinθ = -1/2 sinθ - 2 = 0sinθ = 2it is not possible

sinθ = - sin(π/6)

sinθ = sin(-π/6)

[satisfies the condition sinθ = sinα]

So, α = θ = -π/6.

General angle for sin θ :

θ = nπ + (-1)ⁿ  α    [n ∈ z ]

θ = nπ + (-1)ⁿ (-π/6)

For example

if we put n = 1, we will get

θ = π + (-1)¹ (-π/6) ==> π -π/6 ==> 5π/6

Hence θ = -π/6, 5π/6, ..........

Example 2 :

Solve : 2tanθ - cotθ = -1

Solution :

2tanθ - cotθ = -1

2tanθ - (1/tanθ) = -1

(2tan2θ - 1)/tanθ = -1

Multiply by tanθ on both sides.

(2tan2θ - 1) = -tanθ

2tan2θ  + tanθ - 1 = 0

Let x = tanθ

2x2 + x - 1 = 0

2x2 + 2x - x - 1 = 0

2x(x + 1) - 1(x + 1) = 0

(2x - 1)(x + 1) = 0

 2x - 1 = 02tanθ - 1 = 0Add 1 on both sides2tanθ = 1Divide by 2 on both sidestanθ = 1/2 x + 1 = 0Subtract 1 on both sidesx + 1 - 1 = 0 - 1x = -1tanθ = -1

tanθ = -1

tanθ = -tan(π/4)

tanθ = tan(-π/4)

[satisfies the condition tanθ = tanα]

So, α = θ = -π/4

General angle for tanθ :

θ = nπ + (-1)ⁿ  α    [n ∈ z ]

θ = nπ -(π/4)

tanθ = 1/2

θ = tan-1(1/2)

θ = nπ + (-1)ⁿ  α    [n ∈ z ]

θ = nπ + tan-1(1/2)

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