Lesson HOW TO - Solve Trigonometric equations

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This Lesson (HOW TO - Solve Trigonometric equations) was created by by longjonsilver(2297)

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About longjonsilver: I have a new job in September, teaching

Introduction

The solution of trigonometric equations is one topic that students have particular problems with. There are a few reasons for this:

1. there is usually a simplify part first that requires use of some TRIG identities.
2. there is the use of RADIANS rather than degrees, for which some students are not at ease with.
3. there is the repetitive aspect of TRIG functions that students find bewildering.

All in all, a potentially daunting topic. To be honest though, there is nothing that you need fear here, so long as you take it methodically and slowly. Also, you have to understand what it is you are doing.

Solving the TRIG Equation
Of the 3 topics listed above, I am concentrating on part 3, here in this Lesson. I may well write a Lesson on RADIAN measure at a later date.

First thing, when solving a TRIG equation, is to understand or accept that each of SINE, COSINE and TANGENT have 2 angles that will satisfy the given equation within any 360 degree range. This applies only when we have things like sin(x) or tan(x) + 2 for example.

If we have multiples, such as cos(2x), then we also multiply up the possible number of solutions. In "general", as a rule of thumb, expect the following, within any 360 degree range:

sin(x) --> 2 solutions
sin(2x) --> 4 solutions
sin(3x) --> 6 solutions
sin(4x) --> 8 solutions
sin(5x) --> 10 solutions etc

I shall concentrate here on simple equations, using just (x)... so 2 answers are expected.

Basics
This is difficult to explain, with no real picture to use, so bear with me.

Imagine a set of axes with the 4 quadrants being (in an anti-clockwise manner) 1 to 4 starting with the top right one.

Imagine a line from the origin and lieing on the positive x-axis. This is defined as zero. Now imagine the line turning anticlockwise. When it reaches the positive y-axis, the angle is 90 degrees.

The line carries on rotating, to rest on the negative x-axis... the angle is now 180 degrees.
The line then carries on again, to rest against the negative y-axis... the angle is now 270 degrees

Finally the line completes one revolution and reaches the positive x-direction again. This is 360 degrees (equivalent to 0 degrees).

So, thinking about an angle of say 15 degrees, this is identical in TRIG terms to one of 360+15 --> 375 degrees. Etc.

QUADRANT PROPERTIES
Looking at a set of axes:

Quadrant 1 is the top right one.
Quadrant 2 is the top left one.
Quadrant 3 is the bottom left one.
Quadrant 4 is the bottom right one.

In Quadrant 1 SIN, TAN, COS are ALL +ve
In Quadrant 2 SIN is +ve
In Quadrant 3 TAN is +ve
In Quadrant 4 COS is +ve

eg sin30 = 1/2 and sin150 = 1/2
but sin210 = -1/2 and sin330 = -1/2

what about cos240? well, this angle is in the third quadrant, where cos is negative, so whatever the numerical value, it will be negative, at least! Put it in your calculator...answer is -1/2.

The issues come when you are asked something like "Solve sinx = 1/2 for 0 < x < 360

If you put this into your calculator, you will get 30 degrees as the answer but what about the 150 degree answer too? How do we find that? That is what i am building up to.

Look at the following lines:

{{{graph(200, 200, -6, 6, 3, -3, x/2, -x/2 )}}}

This is the crux of this Lesson... think about the angle each line makes with the x-axis. So in the second quadrant, we have the triangle and the angle between the line and the -ve x-axis is 30 degrees. However, I said earlier that in my view of angles here, angles are measure from the zero of the +ve x-direction, so the line in reality makes an angle of 180-30 --> 150. Look on your calculator at sin(150). this is 1/2.

NEGATIVE questions
If the question asked "Solve sinx=-1/2", then by my method, we would do the following:

1. work out the Quadrant 1 angle, ie solve sinx=1/2 --> this gives 30 degrees.
2. Now, for sin to be negative, we need the quadrant 3 and quadrant 4 angles.

So, these are 180+30 and 360-30. In other words 210 and 330. Find the sin of both of these..you get -1/2.

Done!
That is all there is.

So, examples

EXAMPLES
Q: Solve tanA = 0.4877 for 0 < A < 360
A: The solution is as follows:

1. Find the quadrant1 solution using your calculator, as A = sin^(-1)(0.4877) --> A = 26 degrees, to 2 significant figures).

2. What about the "other" angle that will satisfy this equation? Well, for tan to be positive, the angle must be in Quadrant3. Visualising the third quadrant, with the triangle with an angle of 26 degrees there, this is really an angle of 180+26 --> 206 degrees.

3. check... tan206 is 0.4877

Done!

Q: Solve cosA = -0.5592 for 0 < A < 360
A: The solution is as follows:

1. Find the quadrant1 solution (the POSITIVE VERSION) using your calculator. In other words find angle A = cos^(-1)(0.5592) --> A = 56 degrees.

2. The question wants the angle to give a negative answer, so for COS, this means the angle has to be in Quadrant2 and Quadrant3.

Quadrant2 answer: 180-56 = 124
Quadrant3 answer: 180+56 = 236

3. Check... cos124 = cos236 = -0.5592

Done!

Summary
Hopefully, that is straight-forward? Apologies in not being able to draw a suitably good diagram showing the triangles in each Quadrant, but i think my descriptions and examples show it.

Final Example
Solve sin(2x) = {{{sqrt(3)/2}}} for 0 < x < 360

Solution:
1. Find the quadrant1 angle. This is 60 degrees.

2. Now, the question wants a +ve sin, so that means also Quadrant2. This angle is 180-60 --> 120.

As we will divide this by 2 to find x, we will therefore need to loop round the axes again, so:

3. quadrant1 again is 360+60 --> 420

4. quadrant2 again is 360+120 --> 480

So, now we have

2x = 60, 120, 420, 480

--> x = 30, 60, 210, 240

All these are within the range 0 < x < 360.

Note: we found all the possible angles (2x) before we divided by 2... what we did NOT do is find 2x=60, then divide by 2 to give 30 and THEN find the other angles.

FINALLY Remember to always check your answers on your calculator!