Trig Equation Solving Examples

Trigonometric equations is shortly called as trig equations.  An equation involving trigonometrical function is called a trigonometrical equation.

cosθ =`1/2` , tanθ = 0, cos2θ − 2sinθ =`1/2 `

There are some examples for trigonometrical equations. To solve these equations we find 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. The solution in which the absolute value of the angle is the least is called principal solution. In this article let us study trig equations solving examples.

Trig Equation Solving Examples:

Let us see sample problems for trig equation solving examples.

General solutions of sin θ = 0 ; cosθ = 0 ; tan θ = 0

Find the principal value of the following:

(i) cosx =`sqrt3/2`

Solution: (i) cosx =`sqrt3/2` > 0

∴ x lies in the first or fourth quadrant. Principal value of x must be in[0, π]. Since cosx is positive the principal value is in the first quadrant

cosx =`sqrt3/2` = cos`pi/6` and `pi/6`

∈ [0, π]∴ The principal value of x is `pi/6` .

(ii) cosθ = −`sqrt3/2` < 0

Since cos θ is negative, θ lies in the second or third quadrant. But the  principal value must be in [0, π] i.e.  Within  1st  or  2nd  quadrant. The principal value is in the 2nd quadrant.

cosθ = −`sqrt3/2` = cos (180° − 30°) = cos150°.

The principal value is θ = 150° =`(5pi)/6` .

Trig Equation Solving Examples:

Projection formula

In any triangle ABC  a = b cos C + c cosB

True with usual notations and it is called projection formula.

Proof:

In triangle ABC, draw AD perpendicular to BC.From the right angled triangles ABD and ADC,

cosB =BD

AB ⇒ BD = AB × cosB

cosC =DC

AC ⇒ DC = AC × cosC

But BC = BD + DC = AB cosB + AC cosC

a = c cosB + b cosC

or a = b cosC + c cosB


Solve : sin2x + sin6x + sin4x = 0

Solution:

sin2x + sin6x + sin4x = 0 or (sin6x + sin2x) + sin4x = 0 or 2sin4x. cos2x + sin4x = 0

sin4x (2 cos2x + 1) = 0

when sin4x = 0 ⇒ 4x = nπ or x =`(npi)/4` ; n ∈ Z

When 2 cos2x + 1 = 0 ⇒ cos 2x =− `1/2`

= − cos`pi/3` = cos (π −π3)= cos`(2pi/3)`

∴ 2x = 2nπ ±`(2pi)/3` or x = n π ±`pi/3`

Hence x =`(npi)/4` or x = nπ ±`pi/3`

; n ∈ Z

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