# How do I solve cos2x sinx cosx

## 4.4 Trigonometric equations

### From online math bridging course 1

 theory Exercises

Content:

• Basic trigonometric equations
• Simple trigonometric equations

Learning goals:

After this section you should be able to:

• Solve basic trigonometric equations.
• Solve trigonometric equations that can be converted to other equations.

Are you still very familiar with the learning objectives from school and do you know exactly how to do the calculations? Then you can do the same with the exams begin (you can find the link in the Student Lounge).

### A - Basic equations

Trigonometric equations can be very complicated and often cannot even be solved analytically. But there are some basic trigonometric equations, such as \ displaystyle \ sin x = a, \ displaystyle \ cos x = a and \ displaystyle \ tan x = a, which have relatively simple solutions.

In the general case, such equations either have no or an infinite number of solutions. But if you have the angle x Somehow limited, there are a finite number of solutions, just as if you are looking for an acute angle, for example.

example 1

Solve the equation \ displaystyle \, \ sin x = \ frac {1} {2}.

We want to find all angles that have the sine \ displaystyle \ tfrac {1} {2}. If we look at the unit circle, we see that there are two such angles \ displaystyle x. So here we have the two angles \ displaystyle 30 ^ \ circ = \ pi / 6 and (by symmetry) \ displaystyle 180 ^ \ circ - 30 ^ \ circ = 150 ^ \ circ, which correspond to the yValue \ displaystyle \ sin x = \ tfrac {1} {2}. These are the only such angles between \ displaystyle 0 and \ displaystyle 2 \ pi.

But since we can add a multiple of \ displaystyle 2 \ pi to an angle without changing the sine, we also have the following solutions:

 \ displaystyle \ begin {cases} x & = \ dfrac {\ pi} {6} + 2n \ pi \ x & = \ dfrac {5 \ pi} {6} + 2n \ pi \ end {cases}

where \ displaystyle n is any integer. This is called the general solution to the equation.

If one looks at the graph of \ displaystyle y = \ sin x, one also sees that the equation has an infinite number of solutions, since the curves \ displaystyle y = \ sin x and \ displaystyle y = \ tfrac {1} {2} are infinitely many Have interfaces. Example 2

Solve the equation \ displaystyle \, \ cos x = \ frac {1} {2}.

We consider the unit circle. We know that the cosine of \ displaystyle \ pi / 3 \ displaystyle \ tfrac {1} {2}. If we look at the unit circle, we see that the angle \ displaystyle - \ pi / 3 also has the cosine \ displaystyle \ tfrac {1} {2}. If we add a multiple of \ displaystyle 2 \ pi to these angles, we get the general solution,

 \ displaystyle x = \ pm \ pi / 3 + n \ cdot 2 \ pi \, \ mbox {,}

where \ displaystyle n is any integer.

Example 3

Solve the equation \ displaystyle \, \ tan x = \ sqrt {3}.

We know beforehand that the angle \ displaystyle x = \ pi / 3 satisfies the equation.

If we look at the unit circle, we see that every half turn of the angle has the same slope as the angle and therefore the same tangent. So we get the general solution by adding \ displaystyle \ pi several times to the solution. So we get the solutions \ displaystyle \ pi / 3, \ displaystyle \ pi / 3 + \ pi, \ displaystyle \ pi / 3 + \ pi + \ pi etc. The general solution of the equation is therefore

 \ displaystyle x = \ pi / 3 + n \ cdot \ pi \, \ mbox {,}

where \ displaystyle n is any integer.

### B - More complicated equations

We'll give some examples of more complicated trigonometric equations here.

Some trigonometric equations can be simplified using the trigonometric identities.

Example 4

Solve the equation \ displaystyle \, \ cos 2x - 4 \ cos x + 3 = 0.

We use \ displaystyle \ cos 2x = 2 \ cos ^ 2 \! X - 1 and get

 \ displaystyle (2 \ cos ^ 2 \! x - 1) - 4 \ cos x + 3 = 0 \, \ mbox {,}

By dividing by 2 we get

 \ displaystyle \ cos ^ 2 \! x - 2 \ cos x +1 = 0 \, \ mbox {.}

We factor the left side

 \ displaystyle (\ cos x-1) ^ 2 = 0 \, \ mbox {.}

This equation is only fulfilled if \ displaystyle \ cos x = 1. We solve this equation as before and is the general solution

Example 5

Solve the equation \ displaystyle \, \ frac {1} {2} \ sin x + 1 - \ cos ^ 2 x = 0.

By the Pythagorean Theorem we get \ displaystyle 1 - \ cos ^ 2 \! X = \ sin ^ 2 \! X and we get

 \ displaystyle \ tfrac {1} {2} \ sin x + \ sin ^ 2 \! x = 0 \, \ mbox {.}

We factor out the factor \ displaystyle \ sin x and get

 \ displaystyle \ sin x \, \ cdot \, \ bigl (\ tfrac {1} {2} + \ sin x \ bigr) = 0 \, \ mbox {.}

So we see that the solutions of the equation must satisfy the equations \ displaystyle \ sin x = 0 or \ displaystyle \ sin x = - \ tfrac {1} {2}. We solve these equations as in example 1. The solutions are

 \ displaystyle \ begin {cases} x & = n \ pi \ x & = - \ pi / 6 + 2n \ pi \ x & = 7 \ pi / 6 + 2n \ pi \ end {cases}

Example 6 Solve the equation \ displaystyle \, \ sin 2x = 4 \ cos x.

Using the double angle function for the cosine, we get

 \ displaystyle 2 \ sin x \, \ cos x - 4 \ cos x = 0 \, \ mbox {.}

If we divide by 2 and factor out the factor \ displaystyle \ cos x, we get

 \ displaystyle \ cos x \, \ cdot \, (\ sin x - 2) = 0 \, \ mbox {.}

So the solutions to this equation must be one of the equations

• \ displaystyle \ cos x = 0 \, \ text {or}
• \ displaystyle \ sin x = 2

fulfill. Since \ displaystyle \ sin x is never greater than 1, the second equation has no solutions. The first equation, however, has the solutions \ displaystyle x = \ pi / 2 + n \ cdot \ pi.

Example 7

Solve the equation \ displaystyle \, 4 \ sin ^ 2 \! X - 4 \ cos x = 1.

We use the law of Pythagoras and replace \ displaystyle \ sin ^ 2 \! X with \ displaystyle 1 - \ cos ^ 2 \! X. So we get

 \ displaystyle \ begin {align *} 4 (1 - \ cos ^ 2 \! x) - 4 \ cos x & = 1 \, \ mbox {,} \ 4 - 4 \ cos ^ 2 \! x - 4 \ cos x & = 1 \, \ mbox {,} \ -4 \ cos ^ 2 \! X - 4 \ cos x + 4 - 1 & = 0 \, \ mbox {,} \ \ cos ^ 2 \! X + \ cos x - \ tfrac {3} {4} & = 0 \, \ mbox {.} \ \ end {align *}

This is a quadratic equation for \ displaystyle \ cos x, with the solutions

 \ displaystyle \ cos x = - \ tfrac {3} {2} \ quad \ text {and} \ quad \ cos x = \ tfrac {1} {2} \, \ mbox {.}

Since \ displaystyle \ cos x is never less than \ displaystyle –1, the first equation has no solution. So the equation just has the same solutions as the equation

 \ displaystyle \ cos x = \ tfrac {1} {2} \, \ mbox {,}

which we solved in example 2.

Any questions about this chapter? Then take a look in the course forum (you can find the link in the Student Lounge) or ask ombTutor via Skype  No more questions? Then go on with that Exercises .

Tips for learning

Diagnostic test and final test

After you've finished the theory and exercises, you should take the diagnostic exam and the final exam. You can find the links to the exams in your "Student Lounge".

Keep the following in mind

Learn the basic trigonometric identities and how they are used to simplify trigonometric expressions.

It is important to be familiar with basic trigonometric equations in order to solve more complicated equations. It is also important to know that these equations have infinite solutions.

Useful websites

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