Trigonometric Equation using Formula

We will learn how to solve trigonometric equation using formula.

Here we will use the following formulas to get the solution of the trigonometric equations.

(a) If sin θ = 0 then θ = nπ, where n = 0, ± 1, ± 2, ± 3, …….

(b) If cos θ = 0 then θ = (2n + 1) $$\frac{π}{2}$$, where n = 0, ± 1, ± 2, ± 3, …….

(c) If cos θ = cos ∝ then θ = 2nπ ± ∝, where n = 0, ± 1, ± 2, ± 3, …….

(d) If sin θ = sin ∝ then θ = n π + (-1) $$^{n}$$ ∝, where n = 0, ± 1, ± 2, ± 3, …….

(e) If a cos θ + b sin θ = c then θ  = 2nπ + ∝ ±  β, where cos β = $$\frac{c}{\sqrt{a^{2} + b^{2}}}$$, cos ∝ = $$\frac{a}{\sqrt{a^{2} + b^{2}}}$$ and sin ∝ = $$\frac{b}{\sqrt{a^{2} + b^{2}}}$$, where n = 0, ± 1, ± 2, ± 3, …….

1. Solve tan x + sec x = √3. Also find values of x between 0° and 360°.

Solution:

tan x + sec x = √3

⇒ $$\frac{sin x}{cos x}$$ + $$\frac{1}{cos x}$$ = √3, where cos x ≠ 0

⇒ sin x + 1 = √3 cos x

⇒ √3 cos x - sin x = 1,

This trigonometric equation is of the form a cos θ + b sin θ = c where a = √3, b = -1 and c = 1.

⇒ Now dividing both sides by $$\sqrt{(\sqrt{3})^{2} + (1)^{2}}$$

⇒ $$\frac{√3}{2}$$ cos x - $$\frac{1}{2}$$sin x = $$\frac{1}{2}$$

⇒ cos x cos $$\frac{π}{4}$$ – sin x sin $$\frac{π}{6}$$ = cos $$\frac{π}{3}$$

⇒ cos (x + $$\frac{π}{6}$$) = cos $$\frac{π}{3}$$

⇒ x + $$\frac{π}{6}$$ = 2nπ ± $$\frac{π}{3}$$, where n = 0, ± 1, ± 2, ± 3, …….

⇒ x = 2nπ ± $$\frac{π}{3}$$ - $$\frac{π}{6}$$, where n = 0, ± 1, ± 2, ± 3, …….

When we take minus sign with $$\frac{π}{3}$$, we get

x = 2nπ - $$\frac{π}{3}$$ - $$\frac{π}{6}$$

⇒ x = 2nπ - $$\frac{π}{2}$$, so that cos x = cos (2nπ - $$\frac{π}{2}$$) = cos $$\frac{π}{2}$$ = 0, which spoils the assumption cos x  ≠ 0 (otherwise the given equation would be meaningless).

So, x = 2nπ + $$\frac{π}{3}$$ - $$\frac{π}{6}$$, where n = 0, ± 1, ± 2, ± 3, …….

⇒ x = 2nπ + $$\frac{π}{6}$$, where, n = 0, ± 1, ± 2, ± 3, ……. is the general

solution of the given equation tan x + sec x = √3.

The only solution between 0° and 360° is x = $$\frac{π}{6}$$ = 30°

2. Find the general solutions of θ which satisfy the equation sec θ = - √2

Solution:

sec θ = -  √2

⇒ cos θ = - $$\frac{1}{√2}$$

⇒ cos θ = - cos $$\frac{π}{4}$$

⇒ cos θ = cos (π - $$\frac{π}{4}$$)

⇒ cos θ = cos $$\frac{3π}{4}$$

⇒ θ = 2nπ ± $$\frac{3π}{4}$$, where, n = 0, ± 1, ± 2, ± 3, …….

Therefore, the general solutions of θ which satisfy the equation sec θ = - √2 is θ = 2nπ ± $$\frac{3π}{4}$$, where, n = 0, ± 1, ± 2, ± 3, …….

3. Solve the equation 2 cos$$^{2}$$ x + 3 sin x = 0

Solution:

2 cos$$^{2}$$ x + 3 sin x = 0

⇒ 2(1 - sin$$^{2}$$ x) + 3 sin x = 0

⇒ 2 – 2 sin$$^{2}$$ x + 3 sin x = 0

⇒ 2 sin$$^{2}$$ x – 3 sin x – 2 = 0

⇒ 2 sin$$^{2}$$ x - 4 sin x + sin x – 2 = 0

⇒ 2 sin x(sin x - 2) + 1(sin – 2) = 0

⇒ (sin x - 2)(2 sin x + 1) = 0

⇒ Either sin x - 2 =0 or 2 sin x + 1 = 0

But sin x – 2 = 0 i.e., sin x = 2, which is not possible.

Now form 2 sin x + 1 = 0 we get

⇒ sin x = -½

⇒ sin x =- sin $$\frac{π}{6}$$

⇒ sin x = sin (π + $$\frac{π}{6}$$)

⇒ sin x = sin $$\frac{7π}{6}$$

⇒ x = nπ + (1)$$^{n}$$$$\frac{7π}{6}$$, where, n = 0, ± 1, ± 2, ± 3, …….

Therefore, the solution for the equation 2 cos$$^{2}$$ x + 3 sin x = 0 is x = nπ + (1)$$^{n}$$$$\frac{7π}{6}$$, where, n = 0, ± 1, ± 2, ± 3, …….

Note: In the above trig equation we observe that there is more than one trigonometric function. So, the identities (sin $$^{2}$$ θ + cos $$^{2}$$ θ = 1) are required to reduce the given equation to a single function.

4. Find the general solutions of cos x + sin x = cos 2x + sin 2x

Solution:

cos x + sin x = cos 2x + sin 2x

⇒cos x - cos 2x - sin 2x + sin x = 0

⇒  (cos x - cos 2x) - (sin 2x - sin x) = 0

⇒  2 sin $$\frac{3x}{2}$$ sin $$\frac{x}{2}$$ - 2 cos $$\frac{3x}{2}$$ sin $$\frac{x}{2}$$ = 0

⇒  sin $$\frac{x}{2}$$  (sin $$\frac{3x}{2}$$ - cos $$\frac{3x}{2}$$) = 0
Therefore, either, sin $$\frac{x}{2}$$ = 0

⇒ $$\frac{x}{2}$$= nπ

⇒ x = 2nπ

or, sin $$\frac{3x}{2}$$ -  cos $$\frac{3x}{2}$$ = 0

⇒ sin $$\frac{3x}{2}$$ = cos $$\frac{3x}{2}$$

⇒ tan $$\frac{3x}{2}$$ = 1

⇒ tan $$\frac{3x}{2}$$ = tan $$\frac{π}{4}$$

⇒ $$\frac{3x}{2}$$= nπ + $$\frac{π}{4}$$

⇒ x = $$\frac{1}{3}$$ (2nπ + $$\frac{π}{2}$$) = (4n + 1)$$\frac{π}{6}$$

Therefore, the general solutions of cos x + sin x = cos 2x + sin 2x are x = 2nπ and x = (4n+1)$$\frac{π}{6}$$, Where, n = 0, ±1, ±2, …………………..

5. Find the general solutions of sin 4x cos 2x = cos 5x sin x

Solution:

sin 4x cos 2x = cos 5x sin x

⇒ 2 sin 4x cos 2x = 2 cos 5x sin x

⇒ sin 6x + sin 2x = sin 6x - sin 4x

⇒ sin 2x + sin 4x =0

⇒ 2sin 3x cos x =0

Therefore, either, sin 3x = 0 or, cos x = 0

i.e., 3x = nπ or, x = (2n + 1)$$\frac{π}{6}$$

⇒ x = $$\frac{nπ}{3}$$ or,  x = (2n + 1)$$\frac{π}{6}$$

Therefore, the general solutions of sin 4x cos 2x = cos 5x sin x are $$\frac{nπ}{3}$$ and x = (2n + 1)$$\frac{π}{6}$$

Trigonometric Equations

Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.

Recent Articles

1. Relation between Diameter Radius and Circumference |Problems |Examples

Apr 22, 24 05:19 PM

Relation between diameter radius and circumference are discussed here. Relation between Diameter and Radius: What is the relation between diameter and radius? Solution: Diameter of a circle is twice

2. Circle Math | Terms Related to the Circle | Symbol of Circle O | Math

Apr 22, 24 01:35 PM

In circle math the terms related to the circle are discussed here. A circle is such a closed curve whose every point is equidistant from a fixed point called its centre. The symbol of circle is O. We…

3. Preschool Math Activities | Colorful Preschool Worksheets | Lesson

Apr 21, 24 10:57 AM

Preschool math activities are designed to help the preschoolers to recognize the numbers and the beginning of counting. We believe that young children learn through play and from engaging

4. Months of the Year | List of 12 Months of the Year |Jan, Feb, Mar, Apr

Apr 20, 24 05:39 PM

There are 12 months in a year. The months are January, February, march, April, May, June, July, August, September, October, November and December. The year begins with the January month. December is t…