Trigonometric Equations 1
This chapter explores trigonometric equations. This is the first part of this part. The chapter covers solving harder trigonometric equations of the type for example 2x = 0.4 or cos (x – π/3) = √2. Before attempting this lesson you must have prior knowledge of simple trigonometric equations such as sinx = 0.4
Solving simple equations
The following is a simple equation to start with. Suppose we had to solve;
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Below is the sine graph.
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Above on the graph we can see that for 0.72 the graph gives 46.1°. We’re going to get all the angle for -360° ≤ x ≤ 360°.
Below we can see all the angles even in the opposite angles.
[IMAGE]
So the angles we have managed to find in the point angles were;
[IMAGE]
We need to find the two negative angles which we can see on the graph above. There is a method to do that. The first one is;
[IMAGE]
…and the next one is;
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So the solutions for the question are;
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Solving cos 2x equations
In the following we shall look at cos x. Suppose we had to solve;
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… in the range -180° ≤ x ≤ 180°
Let us first think about cos2x = 0.81 as cosx = 0.81. Now we need to look for the solutions from -360° 50 360° because the question asks for -180° ≤ x ≤ 180°. So we must list all solutions between;
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First let us get an accurate angle for cos-1 (0.81) that is;
[IMAGE]
So now we know the first solution we write;
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The next angle is;
[IMAGE]
Now we can add the next angle to the solutions;
[IMAGE]
Since the graph is symmetrical the other solutions are -324.1° and -35.9°. So now we have;
[IMAGE]
To get x we simply just divide all the angles by 2 so we have;
[IMAGE]
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We managed to get all the angles in the range of -180° and 180°
You can get any of the angles and check whether the answer is right.
…for example let us foe 162° in the original equation.
[IMAGE]
…which is the same as the original equation.
Harder sine
Here is a sine version that we’re going to try and solce. Suppose we wanted to solve;
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…in the range of -180° ≤ x ≤ 180°
First we simplify the equation by dividing by 2 to get rid of the 2. That gives us.
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Now we can sketch the graph os sinx as shown below.
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We find the first angle by using;
[IMAGE]
So now we know the first angle is 17.5 we write;
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The other negative angle is;
[IMAGE]
…now we have;
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There are two other positive angles that we need to find;
[IMAGE]
[IMAGE]
So now we have;
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Remember the angles are for sinx = -0.3 we still need to solve for;
[IMAGE]
…to do that we have to multiply by 3 and divide by 5 so we have;
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…that gives;
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Notice that the last angle is out of range so we remove it that leaves;
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…we can use one of the angle to check where this is true;
[IMAGE]
…which is very similar to the original equation.
Using radians
This time we shall use radians. Suppose we had to solve the following as an example;
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…in the range 0 ≤ x ≤ 2π
Below is the graph in radians
[IMAGE]
…if we were solving cos θ = 1/√2 have been;
[IMAGE]
45° in radians is π/4. This is the first solution. The other positive solution is;
[IMAGE]
We don’t need the negative solutions here because they are out og range, the range asked is 0 ≤ x ≤ 2π. So we have;
[IMAGE]
To find x we need to multiply by 4 and divide by 3 so we have;
[IMAGE]
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The last radian is out of range so we have one radian only;
[IMAGE]
We can check it to see whether this is correct.
[IMAGE]
…this is similar to the original equation.
Remember to always set your calculator to radians instead of degrees.
Solving sin (T – π/3)
In the following example we shall look at solving equations of the form sin (T – π/3). Suppose we wanted to solve;
[IMAGE]
…in the range -π ≤ x ≤ π and giving the answers in radians
We handle this example the same way we did for the previous example we solve for;
[IMAGE]
…and then add on π/3
To find the first radian we use;
[IMAGE]
The other angle is;
[IMAGE]
There are no other angles in the range you could sketch a graph is you’re not sure. So we have;
[IMAGE]
[IMAGE]
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To prove it we can check by using one of the radians in the original equation, for example;
[IMAGE]
Example
Here is another example, suppose we wanted to solve for the following equation;
[IMAGE]
…in the range 0 ≤ θ ≤ 2π
The equation involves;
[IMAGE]
…and we only need to consider in the range of 0 ≤ θ ≤ π
We first do;
[IMAGE]
To find the other radian we;
[IMAGE]
The second radian is out range so we have;
[IMAGE]
…we rearrange to get;
[IMAGE]
…that gives…
[IMAGE]
[IMAGE]
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…so we have…
[IMAGE]Trigonometric Equations 1
This chapter explores trigonometric equations. This is the first part of this part. The chapter covers solving harder trigonometric equations of the type for example 2x = 0.4 or cos (x – π/3) = √2. Before attempting this lesson you must have prior knowledge of simple trigonometric equations such as sinx = 0.4
Solving simple equations
The following is a simple equation to start with. Suppose we had to solve;
[IMAGE]
Below is the sine graph.
[IMAGE]
Above on the graph we can see that for 0.72 the graph gives 46.1°. We’re going to get all the angle for -360° ≤ x ≤ 360°.
Below we can see all the angles even in the opposite angles.
[IMAGE]
So the angles we have managed to find in the point angles were;
[IMAGE]
We need to find the two negative angles which we can see on the graph above. There is a method to do that. The first one is;
[IMAGE]
…and the next one is;
[IMAGE]
So the solutions for the question are;
[IMAGE]
Solving cos 2x equations
In the following we shall look at cos x. Suppose we had to solve;
[IMAGE]
… in the range -180° ≤ x ≤ 180°
Let us first think about cos2x = 0.81 as cosx = 0.81. Now we need to look for the solutions from -360° 50 360° because the question asks for -180° ≤ x ≤ 180°. So we must list all solutions between;
[IMAGE]
First let us get an accurate angle for cos-1 (0.81) that is;
[IMAGE]
So now we know the first solution we write;
[IMAGE]
The next angle is;
[IMAGE]
Now we can add the next angle to the solutions;
[IMAGE]
Since the graph is symmetrical the other solutions are -324.1° and -35.9°. So now we have;
[IMAGE]
To get x we simply just divide all the angles by 2 so we have;
[IMAGE]
[IMAGE]
We managed to get all the angles in the range of -180° and 180°
You can get any of the angles and check whether the answer is right.
…for example let us foe 162° in the original equation.
[IMAGE]
…which is the same as the original equation.
Harder sine
Here is a sine version that we’re going to try and solce. Suppose we wanted to solve;
[IMAGE]
…in the range of -180° ≤ x ≤ 180°
First we simplify the equation by dividing by 2 to get rid of the 2. That gives us.
[IMAGE]
Now we can sketch the graph os sinx as shown below.
[IMAGE]
We find the first angle by using;
[IMAGE]
So now we know the first angle is 17.5 we write;
[IMAGE]
The other negative angle is;
[IMAGE]
…now we have;
[IMAGE]
There are two other positive angles that we need to find;
[IMAGE]
[IMAGE]
So now we have;
[IMAGE]
Remember the angles are for sinx = -0.3 we still need to solve for;
[IMAGE]
…to do that we have to multiply by 3 and divide by 5 so we have;
[IMAGE]
[IMAGE]
…that gives;
[IMAGE]
Notice that the last angle is out of range so we remove it that leaves;
[IMAGE]
…we can use one of the angle to check where this is true;
[IMAGE]
…which is very similar to the original equation.
Using radians
This time we shall use radians. Suppose we had to solve the following as an example;
[IMAGE]
…in the range 0 ≤ x ≤ 2π
Below is the graph in radians
[IMAGE]
…if we were solving cos θ = 1/√2 have been;
[IMAGE]
45° in radians is π/4. This is the first solution. The other positive solution is;
[IMAGE]
We don’t need the negative solutions here because they are out og range, the range asked is 0 ≤ x ≤ 2π. So we have;
[IMAGE]
To find x we need to multiply by 4 and divide by 3 so we have;
[IMAGE]
[IMAGE]
The last radian is out of range so we have one radian only;
[IMAGE]
We can check it to see whether this is correct.
[IMAGE]
…this is similar to the original equation.
Remember to always set your calculator to radians instead of degrees.
Solving sin (T – π/3)
In the following example we shall look at solving equations of the form sin (T – π/3). Suppose we wanted to solve;
[IMAGE]
…in the range -π ≤ x ≤ π and giving the answers in radians
We handle this example the same way we did for the previous example we solve for;
[IMAGE]
…and then add on π/3
To find the first radian we use;
[IMAGE]
The other angle is;
[IMAGE]
There are no other angles in the range you could sketch a graph is you’re not sure. So we have;
[IMAGE]
[IMAGE]
[IMAGE]
To prove it we can check by using one of the radians in the original equation, for example;
[IMAGE]
Example
Here is another example, suppose we wanted to solve for the following equation;
[IMAGE]
…in the range 0 ≤ θ ≤ 2π
The equation involves;
[IMAGE]
…and we only need to consider in the range of 0 ≤ θ ≤ π
We first do;
[IMAGE]
To find the other radian we;
[IMAGE]
The second radian is out range so we have;
[IMAGE]
…we rearrange to get;
[IMAGE]
…that gives…
[IMAGE]
[IMAGE]
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…so we have…
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