14.3 - Trigonometric equations

Before reading this section you may find it helpful to review the following topics: A trigonometric equation is one in which the unknown to be solved for is an angle (call it θ) and that angle is in the argument of a trigonometric function such as sin, cos or tan. A trigonometric equation always has an infinite number of solutions, but it is customary to list only those angles between 0° and 360°.

Only certain types of trigonometric equations can be solved using algebra. We will study the following types:
•   θ occurs only once. Example Go to topic
•   θ occurs several times but always as the argument of the same trigonometric function. Example Go to topic
•   the equation has different trigonometric functions but they all have the same argument, θ. Example Go to topic
•   the equation contains trigonometric functions with different arguments. Example Go to topic


Trigonometric equations in which θ occurs only once

If the unknown angle appears only once, then follow the same procedure as for any other equation where the unknown occurs just once, and invert the operations that were applied to it in the reverse order in which they were applied.



Example:   Solve the equation sin(θ) = ½ for all angles θ between 0° and 360°.



Note: You can use the CAST method to derive the second angle, 150°.



Example: Solve the equation 5 sin(θ − 40°) − 2 = 0 for all angles θ between 0° and 360°.



Note: You can use the CAST method in the second-to-last step to derive the “other” angle, 156.4°. Make sure that you add the 40° last.


Trigonometric equations in which θ occurs several times but always as the argument of the same trigonometric function

If the unknown angle appears more than once, then do the following: A useful trick is to use an alias. Suppose that sin (θ) appears several times in the equation. Then replace all occurrences of sin (θ) with, say Q. This may turn the equation into a linear or quadratic or some other simpler type of equation in the variable Q. Solve this equation first for Q and then substitute back sin (θ) for Q.



Example: Solve the equation 2 sin2(θ) = sin(θ) for all angles θ between 0° and 360°.



Replace this equation by two simpler equations which are solved separately:
Note: You could use an alias to solve this equation. If you replace all occurrences of sin (θ) with Q then this becomes a polynomial equation in the variable Q:
Q 2 = Q.
Solve this equation first for Q:
Q (2 Q − 1) = 0     →     Q = 0   and   Q = ½.
Then substitute back sin (θ) for Q:
sin(θ) = 0   and sin(θ) = ½.
Now solve these quations for θ as above.




Example: Solve the equation 2 sin2(3 θ) = sin(3 θ) for all angles θ between 0° and 360°. Note that this problem is almost identical to the previous one - the only difference is that the argument of the sin function is 3 θ rather than θ.



Replace this equation by two simpler equations which are solved separately:
Note: You can use the CAST method to derive the angles 180°, 360°, and 150°. The other angles are gotten by going around the unit circle multiple times.



Trigonometric equations having different trigonometric functions but all with the same argument

These strategies may help:


Example: Solve the equation 2 csc(θ) − cot(θ) = tan(θ) for all angles θ between 0° and 360°.





Equations containing trigonometric functions with different arguments

If the equation contains trigonometric functions with different arguments (angles) then use the appropriate trigonometric identity to express everything in terms of a single angle. (For example if one angle is twice another angle then use a double angle identity to get rid of the double angle. Or if one angle is another angle plus some difference then use a sum or difference of two angles identity to get rid of the first angle.) Then proceed as before.





Example: The two vectors V1 and V2 point in directions that are 65° apart. Find the value of the angle θ that causes the vertical components of the two vectors (the dotted lines) to be equal in magnitude.

Solution: The length of the vertical component of vector V2 is 5 cos(θ). The length of the vertical component of vector V1 is 3 cos(115° − θ). Using the fact that they are supposed to be equal gives us our starting equation:
5 cos(θ) = 3 cos(115° − θ).
Use trigonometric identity (3b) with the lower sign to separate angle 115° from angle θ on the RHS of the equation.
5 cos(θ) = 3 cos(115°) · cos(θ) + 3 sin(115°) · sin(θ)
Evaluate the sin and cos of 115°. Notice that now the only angle remaining is θ:
5 cos(θ) = −1.268 cos(θ) + 2.719 sin(θ)

Collect terms:
6.268 cos(θ) = 2.719 sin(θ)
Divide both sides by 2.719 cos(θ) and use the basic trigonometric identity to turn sin/cos into tan:
tan(θ) = 2.305

θ = tan−1(2.305) = 66.5° or 246.5°

From the diagram above we see that the angle we want is θ = 66.5°. The other solution corresponds to having the vectors rotated by 180° and being in quadrants 3 and 4.





Example: The following type of problem arises often in electrical technology. Given two sinusoidal waveforms
y1 = 4 sin(θ)   and   y2 = 3 cos(θ + 40°).
Find all angles θ between 0 and 2π radians for which their sum equals 2.

Solution: We must solve the trigonometric equation
4 sin(θ) + 3 cos(θ + 40°) = 2.
Because one angle is 40° more than the other, we will use the sum of angles identity (3b) in the form
cos(θ + 40°) = cos(θ) · cos(40°) − sin(θ) · sin(40°)
to split up the two angles, θ and 40°. This gives
4 sin(θ) + 3 cos(θ) · cos(40°) − 3 sin(θ) · sin(40°) = 2,   or

(4 − 3 sin(40°)) · sin(θ) + 3 cos(40°) · cos(θ) = 2,   or

2.072 sin(θ) + 2.298 cos(θ) = 2.
This looks similar to the previous example but we have a “2” on the right-hand-side so we must use a different method to solve it. In the section on trigonometric identities we proved the following identity, which states that a sine and a cosine waveform can always be added to give a sine waveform with a phase shift:
, where

We can use this identity to replace the two occurrences of θ in our equation by one. In our case A = 2.072 and B = 2.298, so we get C = 3.094 and φ = 47.97° = 0.8372 radians, and our equation becomes
3.094 sin(θ + 0.8372) = 2.
Now θ occurs just once and we simply isolate it:
sin(θ + 0.8372) = 0.6464

θ + 0.8372 = sin−1(0.6464) = 0.7029,   π − 0.7029   and   0.7029 + 2 π

Of the three angles listed on the right-hand-side, The last step to isolate θ is to subtract 0.8372 from both sides of the equation:
θ = −0.134,   1.602   and   6.149


The graph to the right shows the two waveforms y1 and y2 (the green and blue curves) as well as their sum (the red curve). The red dots in the graph are the locations where the sum equals 2. To answer the question we drop the first solution because it is not between 0 and 2π and keep the other two.




Example: Voltage v (the dark line) is the sum of two
sinusoidal waves (the two light lines) and is given by the equation:
v = 100 sin(1 t) + 50 cos(2 t).
Determine the times t when v = 0 in the cycle shown (indicated by the black dots).

(Note that the angular velocities of the sinusoidal waves are 1 radian/sec and 2 radians/sec and that time t is in seconds, so the angles 1 t and 2 t are in radians.)

Solution: Because one angle is twice the other, use the double angle identity in the form:
cos(2 t) = 1 − 2 sin2(t)
to get rid of the angle 2 t in favor of the angle t:
0 = 100 sin(t) + 50·{1 − 2 sin2(t)}
Distribute, transpose everything to the left side of the equation and divide by 50:
100 sin2(t) − 100 sin(t) − 50 = 0
2 sin2(t) − 2 sin(t) − 1 = 0
This is a quadratic equation in sin(t). (In fact replacing sin(t) with the alias Q would turn it into the quadratic equation 2 Q 2 − 2 Q − 1 = 0.) Plugging a = 2, b = −2 and c = −1 into the quadratic formula we get:
It is impossible for the sin of anything to equal 1.366. Thus we are left with:
sin(t) = −0.366

t = sin−1(−0.366)

Putting the calculator in radian mode gives t = −0.3747. However the unit circle picture to the right shows that the values we require to answer this question are t1 = 3.52 and t2 = 5.91 seconds.





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