Trigonometric Equations

ObjectivesObjetivos

At the end of this lesson, one should be able to:

• Solve trigonometric equations in the interval [0, 2$\pi$ ) .
• Find the general solution of a trigonometric equation.

Introduction

A trigonometric equation is an equation that contains trigonometric expressions and is solved using similar techniques like the ones used in solving algebra equations, in which that the solutions represent angles.

For example the following are trigonometric equations:

1. $2\sin x=1$
2. ${\sin }^{2}x+{\cos }^{2}x=1$
3. ${\tan }^{2}2x-1=0$

If the measure of the gradients is not specified, we will work in radians.

Solving Trigonometric Equations

We will solve the trigonometric equations in the interval [0, 2 $\pi$ ) and also in the general form.

Example 1

Find the solution of the equation, $\sin x=\frac{1}{2}$

• In the interval [0, 2 $\pi$ ).

• In all the real numbers

$$ Solution:

If $\sin x=\frac{1}{2}$ then the angle of reference for x is x = $\frac{\pi }{6}$ , which can be found in the quadrant I; if we consider other solutions of the equation we will see that $x=\pi -\frac{\pi }{6}=\frac{\mathrm{5\; \pi }}{6}$ , located in the quadrant II, is the other solution in the interval [0, 2 $\pi$ ), a continuation is shown graphically.

Like the function sin has a period 2 $\pi$ , we can obtain all the solutions, summing the mutiples of 2 $\pi$ to $\pi /6\; y$ to 5 π /6, and then we have:

$x=\frac{\pi }{6}+2\mathrm{\pi n}$ ,   $x=\frac{5\pi }{6}+2\mathrm{\pi n}$

for all of n.

Example 2

Find all the solutions of the equation, $2\cos x-1=0$

Solution:

$2\cos x-1=0$

2$\cos x=$1

$\cos x=\frac{1}{2}$

Analogously with the previous example we have the graph

Observing the image we find that the angles that solve the equation in [0, 2 $\pi$ ), are:

$x=\frac{\pi }{3}$ ,   $x=\frac{5\pi }{3}$ , solutions in the I and IV quadrant respectively.

Like the function cos has a period 2 $\pi$ , we sum the multiples of this number and the anterior solutions.

$x=\frac{\pi }{3}+2\mathrm{\pi n}$ ,   $x=\frac{5\pi }{3}+2\mathrm{\pi n}$

for all of n.

Solving Trigonometric Equations that involve double angles

We solve the trigonometric equations whose trigonometric functions wrap double angles, in the interval [0, 2 $\pi$ ) and also in the general form.

Example 3

Find all the solutions of the equation, $\cos 2x=0$

Solution:

We know that if $\cos \theta =0$ , then:

$\theta =\frac{\pi }{2}+\mathrm{\pi n}$ 

will θ = 2x then $\mathrm{2x}=\frac{\pi }{2}+\mathrm{\pi n}$ , then

$x=\frac{\pi }{4}+\frac{\pi }{2}n$

In gradients, we have: $x={45}^o+{90}^{o}n$

The figure shows the solutions of the trigonometric equations cosine of the double angles.

Solving Trigonometric Equations through Factorization

We use factorization to solve trigonometric equations.

Example 4

Find all the solutions of the equation, $\sin \mathrm{(\theta )}\tan \mathrm{(\theta )}=\sin \theta$

Solution:

$\sin \left(\theta \right)\tan \left(\theta \right)=\sin (\theta )$ gives

$\sin \left(\theta \right)\tan \left(\theta \right)-\sin (\theta )=0$ making one side zero

$\sin \left(\theta \right)(\tan \theta -1)=0$ common factor

$\sin \theta =0$ y $\tan \theta -1=0$ equating every factor to zero

$\sin \theta =0$ y $\tan \theta =1$ solving for sine and tangent

The solutions of the sine are: 0,π,-π,2π,-2π,...

in general if $\sin \theta =0$ , then θ = πn for the whole n.

In another way the tangent function has period π like the solution of $\tan \theta =1$ in the interval (-π/2,π/2) es π/4

in general if $\tan \theta =1$ , then θ=$\frac{\pi }{4}+\mathrm{\pi n}$

Finally the solution of the equation given is:

θ = πn and θ=$\frac{\pi }{4}+\mathrm{\pi n}$ for all of n. Graphically we have:

Example 5

Find all the solutions of the equation,${\mathrm{2\; sin}}^{2}x-\sin x-1=0$

Solution:

The trigonometric equation is the quadrant type $2y^2-y-1$ , that factorizes in (2y+1)(y-1).

So our factorized trigonometric equation is:

$(2\sin x+1)\left(\sin x-1\right)=0$

2$\sin x+$1=0 and sin$x-$1=0

2$\sin x=-1$ and sin$x=1$

$\sin x=-1/2$ and sin$x=$1

in the interval: [0, 2 $\pi$ )

$x=\frac{7\pi }{6},\frac{11\pi }{6},\frac{\pi }{2}$

In general we have:

$x=\frac{7\pi }{6}+2\mathrm{n\pi },\frac{11\pi }{6}+2\mathrm{n\pi },\frac{\pi }{2}+2\mathrm{n\pi }$ , graphically we can observe the solutions.

Example 6

Find all the solutions of the equation, $\cos x+1=\sin x$

Solution:

When you cannot find any identity to reduce the expression or it cannot be factorized, it is useful to elevate some sides of the quadrant equation, but one has to verify the solution that is found.

$\cos x+1=\sin x$

${\left(\cos x+1\right)}^2={(\sin x)}^2$

${\cos }^{2}x+2\cos x+1=1-{\cos }^{2}x$

${\mathrm{2cos}}^{2}x$+2$\cos x=0$

2cos$x(\cos x+1)=$0

$2\cos x=0$ and $\cos x+1=0$

$\cos x=0$ and $\cos x=-1$

x$=\frac{\pi }{2},\frac{3\pi }{2},\pi$

Verification:

si $x=\frac{\pi }{2}$, we have:

$\cos \frac{\pi }{2}+1=\sin \frac{\pi }{2}$

0+1=1 correct.

si $x=\frac{3\pi }{2}$, we have:

$\cos \frac{3\pi }{2}+1=\sin \frac{3\pi }{2}$

0+1=-1 false.

si x = π, we have:

$\cos \pi +1=\sin \pi$

-1+1=0 true

In general: x$=\frac{\pi }{2}+2\mathrm{n\pi },\pi +2\mathrm{n\pi }$

The graph shows the trigonometric equation as the same as zero, clearly it is noted that the roots of the fuction represent the solutions of the trigonometric equation.

To practice exercises about the trigonometric equations click on the following button

Trigonometric Equations

Summary

Now that this lesson has been completed, you should be able to:

• Solve trigonometric equations in the basic interval [0, 2 $\pi$ ).
• Solve trigonometric equations in the general form.