Solving Trigonometric equations

How to solve trigonometric equations.

To solve a trigonometric equation, we use standard algebraic techniques such as collecting like terms and factoring. Our preliminary goal in solving a trigonometric equation is to isolate the trigonometric function in the equation. We can use trigonometric equations to solve a variety of real life problems.

Lets first look at some basic equations in this video.

Example1: Solve sinx +√2 = -sinx

Solution: First we collect like terms.

sinx + sinx +√2 = 0

2sinx +√2 = 0 [ add like terms]

sinx = -√2/2 [isolate sinx ]

x = 5π/4 , 7π/4

First we find all possible solutions in interval [0,2π ], then we add multiples of 2π to each of these solutions to get general form. So its general solution is given as,

$\dpi{120} x=\frac{5\pi }{4}+2n\pi ,\frac{7\pi }{4}+2n\pi$ where n is an integer.

Example2: Solve the given trigonometric equation and write all possible general solutions.

$\dpi{120} {\color{Red} 3sec^{2}x-4=0}$

Solution:

$\dpi{120} 3sec^{2}x-4=0$

$\dpi{120} 3sec^{2}x=4$

$\dpi{120} sec^{2}x=\frac{4}{3}$ [square root both the sides]

$\dpi{120} secx=\pm \sqrt{\frac{4}{3}}$

$\dpi{120} \frac{1}{cosx}= \pm \frac{2}{\sqrt{3}}$

$\dpi{120} cosx = \pm \frac{\sqrt{3}}{2}$

Since cos has a period of [0, 2π] so we find all the solutions in this interval ,taking into account both positive and negative signs. So we get,

x =π/6 , 5 π/6 , 7π/6 , 11π/6

General solution is given as,

$\dpi{120} x=\frac{\pi }{6}+2n\pi , \frac{5\pi }{6}+2n\pi , \frac{7\pi }{6}+2n\pi , \frac{11\pi }{6}+2n\pi$

Where n is an integer.

Solving trigonometric equations given in quadratic form.

Some trigonometric equations are given in quadratic form or they are given in such form which can be converted to quadratic form. Such equations can be solved using factoring by different ways as shown in these following examples.

Example: Find all solutions of the equation in the interval [0,2π].

${\color{Red} 2cos^2(x)-5cosx+2=0}$

Solution:

As this equation is quadratic in nature so we use factoring to get its factors.

To make factoring process easy, we can assume cosx as y and get the factors of

$\dpi{120} 2y^{2}-5y+2 =0$

(2y-1)(y-2) = 0

Plug in back y and we can get the factors as,

(2cosx-1)(cosx-2) = 0

2cosx -1 = 0 cosx -2 = 0

Cosx = 1/2 cosx = 2 [no solution]

Note: Here no solution possible for cosx =2 as 2 lie outside the range of cosine function.

So its solution in the given interval is,

$cosx =\frac{1}{2}$

$x=\frac{\pi }{3}, \frac{5\pi }{3}$

Example : Solve the given trigonometric equation and find its general solutions.

${\color{Red} tan^4(\theta )-13tan^2(\theta )+36 =0}$

Example3: Solve the given trigonometric equation and write all possible general solutions.

$\dpi{120} {\color{Red} cos^{3}x=cosx}$

Solution:

$\dpi{120} cos^{3}x=cosx$

$\dpi{120} cos^{3}x-cosx=0$

$\dpi{120} cosx \left (cos^{2}x-1 \right )=0$ [factored out cosx]

set each factor =0 and solve for x,

cosx =0 $\dpi{120} cos^{2}x-1=0$

x = π/2 , 3π/2 $\dpi{120} cosx=\pm 1$

x = π/2 , 3π/2 x = 0, π

These are the solutions in the interval [0,2π]

To get general solutions we add 2nπ to each primary solution.

x=π/2+2nπ , 3π/2+2nπ ,0+2nπ, π+2nπ

π/2+2nπ , 3π/2+2nπ can be combined to get π/2+nπ and hence general solutions are given as,

x = π/2+nπ , 2nπ , π+2nπ , n∈Z

Solving trigonometric equations by squaring both sides.

Sometimes we are not able to solve trigonometric equations either by factoring or using any identity. In such case we can try squaring both sides and then solving it further as shown in following examples.

Example: Find all solutions of the equation in the interval [0,2π]

cscx + cotx = 1

Solution: In such cases we square both sides and solve .

$\dpi{120} \left ( cscx \right )^{2}=\left ( 1-cotx \right )^{2}$

$\dpi{120} csc^{2}x= 1-2cotx +cot^{2}x$

$\dpi{120} csc^{2}x-cot^{2}x=1-2cotx$

$1= 1-2cotx$

$0=-2cotx$

$0= cotx$

$\frac{\pi }{2}= x$

Example : Solve the given trigonometric equation and write its general solutions. Also check extreneous solutions.

${\color{Red} sin(\theta )-1=cos(\theta )}$

Solving multiple angle trigonometric equations.

Example: Solve the multiple angle equation and write its general solution.

$\dpi{120} {\color{Red} cos\left ( \frac{x}{2} \right )=\frac{\sqrt{2}}{2}}$

Solution:

$\dpi{120} cos\left ( \frac{x}{2} \right )=\frac{\sqrt{2}}{2}$

$\dpi{120} \frac{x}{2}= \frac{\pi }{4}, \frac{7\pi }{4}$ [cos being positive in 1^st and 4^th quadrants]

This is the solution in the interval [0,2π] as cos has a period of 2π.

Its general solution would be given as,

$\dpi{120} \frac{x}{2}=\frac{\pi }{4}+2n\pi , \frac{7\pi }{4}+2n\pi$

Multiplying both sides by 2 to Isolate x we get,

$\dpi{120} x=\frac{\pi }{2}+4n\pi , \frac{7\pi }{2}+4n\pi , n\in Z$

Practice problems:

Solve the given trigonometric equation and write all possible general solutions.

$\dpi{120} cotx cos^{2}x =2cotx$
$\dpi{120} 2sec^{2}x +tan^{2}x-3=0$
Sin(2x) = -√3/2
$\dpi{120} 2sin^{2}x +3cosx-3 =0$

Answers:

$\dpi{120} x=\frac{\pi }{2}+n\pi$
$\dpi{120} x=\frac{\pi }{6}+2n\pi ,\frac{5\pi }{6}+2n\pi ,\frac{7\pi }{6}+2n\pi ,\frac{11\pi }{6}+2n\pi$
$\dpi{120} x=\frac{2\pi }{3}+n\pi , \frac{5\pi }{6}+n\pi$
$\dpi{120} x=2n\pi ,\frac{\pi }{3}+2n\pi ,\frac{5\pi }{3}+2n\pi$

Where n is an integer.

How to solve trigonometric equations.

Solving trigonometric equations given in quadratic form.

Solving trigonometric equations by squaring both sides.

Solving multiple angle trigonometric equations.

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