Solving Trig Equations |
Solving Trig Equations:
**WARNING: You cannot solve trigonometric equations if you don't know the values associated with the unit circle and the 2 right triangles, (45°, 45°, 90° and 30° 60°, 90°).
Note: Sometimes the range of solution values is stated.
In such a case, our answers must correspond to the given range.
We must verify if the required solutions should be expressed in degrees or radians.
If the stated range is 0° < A < 360° , then our answers must be in this range of degrees.
If however the stated range is , then we must answer in radians.
However, if the question asks for solutions in R, we must include all possible values of the solutions. Remember that coterminal angles have the same trig function values, so all must be listed.
To do this, we attach a multiple of or to the solution. We do this by the following: say our answer is . In order to express all the possible angles we simply state the answer
as (the set of integers).
Had our answer been (1st and 3rd quads) we would express the solution as .
Note: with a mix of trig functions, it's usually best to change everything to sine or cosine before solving.
Examples
Let's solve some trig equations.
Note: trig expressions can be factored just like algebraic expressions.
1) Solve 2 sin A cos A = sin A, | 0° < A < 360° |
2 sin A cos A – sin A = 0 | |
sin A (2 cos A – 1) = 0 | |
sin A = 0 | 2 cos A – 1 = 0 |
sin A = 0 at A = 0° or 180° | cos A = ½ at A = 60^{ 0} or 300° |
remember that the trig functions have different signs in the 4 quadrants.
Since cos A > 0 in the 1st and 4th quads, we have to include both angles.
2) Solve tan^{ 2} u + tan u = 0 | |
tan u ( tan u + 1) = 0 | |
tan u = 0 | tan u = –1 (tan u < 0 in 2nd and 4th quads) |
u = 0 or |
.
3) tan^{ 2} A – 2 tan A – 3 = 0 | 0° < A < 360° |
(tan A – 3)(tan A + 1) = 0 | |
tan A = 3 | tan A = – 1 (note: we solved tan A = –1 in #2) |
A = arctan 3 gives us A = 71.57° or 251.57° | A = 135° or A = 315° |
(since tan A > 0 in 1st and 3rd quads) |
Note: for trig values not associated with the unit circle or the two right triangles we know,
we use inverse trig functions to find the solutions, as we did in this example.
Remember that sin A and cos A are restricted to values between –1 and 1. If you get
sin A = – 4, or cos A = 7/5, there are no solutions since both these values are outside the
range of these two trig functions.
4) sin 2A cos A – sin A = 0 | 0° < A < 360° |
2 sin A cos^{ 2} A – sin A = 0 | (using sin 2A = 2 sin A cos A) |
sin A (2 cos^{ 2} A – 1) = 0 | |
sin A = 0 | or cos^{ 2} A = ½ |
A = 0 or A = 180° | A = 45°, 135°, 225°, or 315° |
5) 3 sin^{ 2} u + sin u – 2 = 0 | u ` R |
(3 sin u – 2) (sin u + 1) = 0 | |
sin u = 2/3 | sin u = –1 |
u = arcsin 2/3 | u = arcsin (–1) |
. |
.
.
Practice
Solve these equations in R.
1) 2 cos^{ 2}A + sin A = 1
2) tan A + 3 cot A = 4
3) 3 – 4 sin^{ 2}A = 2 cos^{ 2}A
4) tan^{ 2}A + sec^{ 2}A – 7 = 0
5) cot A – 5 csc A + 3 tan A = 0
.
Solutions
1) 2 cos^{ 2}A + sinA = 1 | 2) tanA + 3 cotA = 4 | ||
2(1 – sin^{ 2}A) + sinA = 1 | tanA + 3/tanA = 4 | ||
1 + sinA – 2sin^{ 2}A = 0 | tan^{ 2}A – 4 tanA + 3 = 0 | ||
(1 + 2sinA)(1 – sinA) = 0 | (tanA – 3)(tanA –1) = 0 | ||
sinA = –½ | sinA = 1 | tanA = 3 | tanA = 1 |
A = arctan 3 | A = arctan 1 = . | ||
3) 3 – 4 sin^{ 2}A = 2 cos^{ 2}A | 4) tan^{ 2}A + sec^{ 2}A – 7 = 0 | ||
3 – 4 sin^{ 2}A = 2(1 – sin^{ 2}A) | tan^{ 2}A + (1 + tan^{ 2}A) – 7 = 0 means tan^{ 2}A = 3 | ||
1 – 2 cos^{ 2}A = 0 | so | ||
so | so |
5) cotA – 5 cscA + 3 tanA = 0 | |
cos^{ 2}A – 5 cosA + 3 sin^{ 2}A = 0 | |
cos^{ 2}A – 5 cosA + 3(1 – cos^{ 2}A) = 0 | |
3 – 5 cosA – 2 cos^{ 2}A = 0 | |
(3 + cosA)(1 – 2 cosA) = 0 | |
cosA = – 3 doesn't exist but cosA = ½ | |
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