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4.4 Trigonometric equations

From Förberedande kurs i matematik 1

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       Theory          Exercises      

Contents:

  • Simple trigonometric equations

Learning outcomes:

After this section, you will have learned how to:

  • Solve the basic equations of trigonometry.
  • Solve trigonometric equations that can be reduced to basic equations.

Basic equations

Trigonometric equations can be very complicated, but there are also many types which can be solved using relatively simple methods. Here we shall start by looking at the most basic trigonometric equations, of the type sinx=a, cosx=a and tanx=a.

These equations usually have an infinite number of solutions, unless the circumstances limit the number of possible solutions (for example, if one is looking for an acute angle).

Example 1

Solve the equation sinx=21.


Our task is to determine all the angles that have a sine equal to 21. The unit circle helps us in this. Note that here the angle is designated as x.

[Image]

The figure illustrates the two points on the circle which have y-coordinate 21, i.e. the two points whose corresponding angles have sine value 21. The first is the standard angle 30=6 and by symmetry the other angle makes 30 with the negative x-axis. Therefore the other angle is 180–30=150 or in radians –6=56. These are the only solutions to the equation sinx=21 between 0 and 2.

However, we can add an arbitrary number of revolutions to these two angles and still get the same value for the sine . Thus all angles x for which sinx=21 are

xx=6+2n=65+2n

where n is an arbitrary integer. This is called the general solution to the equation.

The solutions can also be obtained from the figure below, by looking at where the graph of y=sinx intersects the line y=21.

[Image]

Example 2

Solve the equation cosx=21.


We once again study the unit circle.

[Image]

We know that cosine is 21 for the angle 3. The only other point on the unit circle which has x-coordinate 21 corresponds to the angle −3. Adding an integral number of revolutions to these angles we get the general solution

x=3+n2,

where n is an arbitrary integer.

Example 3

Solve the equation tanx=3 .


One solution to the equation is the standard angle x=3.

If we study the unit circle then we see that the tangent of an angle is equal to the slope of the straight line through the origin which makes an angle x with the positive x-axis .

[Image]

Therefore, we see that the solutions to tanx=3  repeat themselves every half revolution, and so the general solution can be obtained from the solution 3 by adding or subtracting multiples of :

x=3+n,

where n is an arbitrary integer.


Somewhat more complicated equations

Trigonometric equations can vary in many ways, and it is impossible to give a full catalogue of all possible equations. But let us study some examples where we can use our knowledge of solving basic equations.

Some trigonometric equations can be simplified by rewriting them with the help of the trigonometric relationships. This, for example, could lead to a quadratic equation, as in the example below.

Example 4

Solve the equation cos2x–4cosx+3=0.


Rewrite by using the formula cos2x=2cos2x–1, so that

(2cos2x–1)–4cosx+3=0.

Simplifying, we see that

cos2x−2cosx+1=0.

The left-hand side can factorised by using the squaring rule to give

(cosx−1)2=0.

This equation can only be satisfied if cosx=1. The basic equation cosx=1 can be solved in the normal way, and thus the complete solution is

x=2n(n arbitrary integer).

Example 5

Solve the equation 21sinx+1–cos2x=0.


According to the Pythagorean identity sin2x+cos2x=1, i.e. 1–cos2x=sin2x. Thus the equation can be written as

21sinx+sin2x=0.

Factorising out sinx we get

sinx21+sinx=0. 

From this factorised form of the equation, we see that the solutions either have to satisfy sinx=0 or sinx=−21, which are two basic equations of the type sinx=a and can be solved as in Example 1. The solutions turn out to be

xxx=n=−6+2n=76+2n(n  arbitrary integer).

Example 6

Solve the equation sin2x=4cosx.


By rewriting the equation using the formula for double-angles we get

2sinxcosx–4cosx=0.

Dividing both sides by 2 and factorising out cosx, gives

cosx(sinx–2)=0.

The left-hand side can only be zero if one of the factors is zero, and we have reduced the original equation into two basic equations:

  • cosx=0,
  • sinx=2.

However, sinx can never be greater than 1, so the equation sinx=2 has no solutions. That just leaves cosx=0, and using the unit circle we see that the general solution is

x=2+n,

for n an arbitrary integer.

Example 7

Solve the equation 4sin2x–4cosx=1.


Using the Pythagorean identity we can replace sin2x by 1–cos2x. Then

4(1–cos2x) – 4cosx4 – 4cos2x – 4cosx–4cos2x – 4cosx + 4 – 1cos2x+cosx–43=1=1=0=0.

This is a quadratic equation in cosx, which has the solutions

cosx=−23andcosx=21.

Since the value of cosx is between –1 and 1, the equation cosx=−23 has no solutions. That leaves only the basic equation

cosx=21,

which may be solved as in Example 2.


Exercises

Study advice

Basic and final tests

After you have read the text and worked through the exercises, you should do the basic and final tests to pass this section. You can find the link to the tests in your student lounge.


Keep in mind that...

It is a good idea to learn the most common trigonometric formulas (identities) and practice simplifying and manipulating trigonometric expressions.


It is important to be familiar with the basic equations, such as sinx=a, cosx=a or tanx=a (where a is a real number). It is also important to know that these equations typically have infinitely many solutions.


Useful web sites

Experiment with the graph y = a sin b (x-c)

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