Trigonometric Identities and Equations
9.5 Solving Trigonometric Equations
Solving Trigonometric Equations
Now it is time to take what we've learned and apply it to solving equations with trigonometric functions in them. Because trigonometric functions are periodic, these questions will often have a general solution and a more specific one. The general solution to equations like these are infinite sets of solutions.
Example: Solving a trig equation
Find all solutions to the equation $2\sin\theta = -\sqrt{3}$.
::solution Just like solving any equation, we need to isolate our variable. Back in algebra, in most cases it was enough to add, subtract, multiply, and divide to isolate a variable. Sometimes we needed to root, but in these problems we have a new hurdle. Our variable is in a trig function. So we'll need to use our inverse trig functions.
So we found one answer, but we know that our inverse sine only gave us one of the two that could have been. The other angle is over in quadrant III. Our reference angle is $\frac{\pi}{3}$, in quadrant III that would be $\frac{4\pi}{3}$. However, for any solution we have, we can add any multiple of the period for the function to get yet another solution.
The general solution to the problem is
$$\theta_1 = -\frac{\pi}{3} + 2\pi k$$ or $$\theta_2 = \frac{4\pi}{3} + 2\pi k$$
where $k\in\mathbb{Z}$. We can make a table to calculate these values.
| k | $-\frac{\pi}{3} + 2\pi k$ | $\frac{4\pi}{3} + 2\pi k$ |
|---|---|---|
| 0 | $-\frac{\pi}{3}$ | $\frac{4\pi}{3}$ |
| 1 | $\frac{5\pi}{3}$ | $\frac{10\pi}{3}$ |
| 2 | $\frac{11\pi}{3}$ | $\frac{16\pi}{3}$ |
And you can see how this table can go on forever. We'll just give the first few rows as evidence that our solution is indeed general. :::
MyOpenMath: General Solution to a Trig Equation
Usually, the problem comes with an interval on which to give your solutions. A natural interval would be to find all of the solutions on the interval from $0\leq \theta < 2\pi$, but this need not always be the case.
Example: Basic trig equation
Find all the solutions for $2\cos\theta - 3= -5$ on the interval $0\leq \theta < 2\pi$.
::solution First, we'll find the general solution.
This particular problem is right on the edge of the range for cosine's inverse so there isn't a second solution the first time around the circle that we need to worry about. Our general solution is
$$\theta=\pi + 2\pi k$$
Plugging in values for $k$ we see that we only need $\pi$.
| k | $\pi + 2\pi k$ |
|---|---|
| -1 | $-\pi$ |
| 0 | $\pi$ |
| 1 | $3\pi$ |
So our solution set is just {$\pi$}. :::
Example: Basic trig equation
Find all the solutions for $2\sin\theta= \sqrt{2}$ on the interval $0\leq \theta < 2\pi$.
::solution First find the general setup:
Sine is positive in quadrants I and II, with reference angle $\frac{\pi}{4}$. So on one cycle,
$$ \theta=\frac{\pi}{4}\quad\text{or}\quad\theta=\frac{3\pi}{4} $$
Both are in $0\leq\theta<2\pi$, so the solution set is
$$ \{\frac{\pi}{4},\frac{3\pi}{4}\} $$ :::
MyOpenMath: General Solution to a Trig Equation
Example: Involving tangent
Find all the solutions for $3\tan^2\theta + 1 = 4$ on the interval $0\leq \theta < 2\pi$.
Show Solution
Solve for tangent first:
On $0\leq\theta<2\pi$:
- $\tan\theta=1$ at $\theta=\frac{\pi}{4},\frac{5\pi}{4}$
- $\tan\theta=-1$ at $\theta=\frac{3\pi}{4},\frac{7\pi}{4}$
So the solution set is
$$ \{\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\} $$
MyOpenMath: General Solution to a Trig Equation
Always Factor, Never Divide
As a good rule of thumb, when possible you should always factor. Recall back in the land of algebra it was dangerous to solve something like $x^2-x=0$ You might be tempted to divide everything by $x$ and get $x-1=0$ leading you to the solution that $x=1$. But, you don't know if the $x$ you divided by was zero or not. In fact, we missed a zero here. Going back to square one:
We need to keep the same things in mind here and factor when possible. Especially if it is a product of terms set equal to zero.
Example: Factoring for a solution
Find all the solutions for $2\sin\theta\cos\theta=\cos\theta$ on the interval $0\leq \theta < 2\pi$.
Show Solution
Move everything to one side and factor:
So either
$$ \cos\theta=0\quad\text{or}\quad 2\sin\theta-1=0 $$
From $\cos\theta=0$: $\theta=\frac{\pi}{2},\frac{3\pi}{2}$.
From $\sin\theta=\frac{1}{2}$: $\theta=\frac{\pi}{6},\frac{5\pi}{6}$.
So the solution set is
$$ \{\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6},\frac{3\pi}{2}\} $$
MyOpenMath: Trig equation with factoring
Example: Factoring again
Find all the solutions for $2\cos^2\theta=\cos\theta$ on the interval $0\leq \theta < 2\pi$.
Show Solution
Again, set the equation to zero and factor:
So either
$$ \cos\theta=0\quad\text{or}\quad 2\cos\theta-1=0 $$
From $\cos\theta=0$: $\theta=\frac{\pi}{2},\frac{3\pi}{2}$.
From $\cos\theta=\frac{1}{2}$: $\theta=\frac{\pi}{3},\frac{5\pi}{3}$.
So the solution set is
$$ \{\frac{\pi}{3},\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{3}\} $$
MyOpenMath: Trig equation with factoring
Trig equations with altered arguments
When our trigonometric equations start involving more transformations, specifically ones that change the period or phase shift, we can end up having far more (or fewer) solutions on a given interval.
This doesn't really pose an issue for our general solution. But we will need to check more values for $k$ to make sure we've found all solutions.
Example: Solving Equations with shorter periods
Find all the solutions for $2\sin(4\theta)-1=0$ on the interval $0\leq \theta < 2\pi$.
Show Solution
Start by isolating the trig function:
Now solve for the angle $4\theta$:
$$ 4\theta=\frac{\pi}{6}+2\pi k\quad\text{or}\quad 4\theta=\frac{5\pi}{6}+2\pi k $$
Divide by 4:
$$ \theta=\frac{\pi}{24}+\frac{\pi}{2}k\quad\text{or}\quad \theta=\frac{5\pi}{24}+\frac{\pi}{2}k $$
To stay in $0\leq\theta<2\pi$, use $k=0,1,2,3$.
From $\theta=\frac{\pi}{24}+\frac{\pi}{2}k$:
$$ \frac{\pi}{24},\frac{13\pi}{24},\frac{25\pi}{24},\frac{37\pi}{24} $$
From $\theta=\frac{5\pi}{24}+\frac{\pi}{2}k$:
$$ \frac{5\pi}{24},\frac{17\pi}{24},\frac{29\pi}{24},\frac{41\pi}{24} $$
So the solution set is
$$ \{\frac{\pi}{24},\frac{5\pi}{24},\frac{13\pi}{24},\frac{17\pi}{24},\frac{25\pi}{24},\frac{29\pi}{24},\frac{37\pi}{24},\frac{41\pi}{24}\} $$
MyOpenMath: General Solution with different frequencies
Quadratic, but not quite
Occasionally we'll bump into equations that resist our more common factoring techniques.
Example: Quadratic trig equation
$$2\sin^2\theta - 3\sin\theta + 1 = 0$$
Show Solution
But, if you squint this looks kinda like a quadratic equation $ax^2+bx+c=0$. It turns out, that we can solve it using the same methods that one would use to solve a quadratic. Lets make a quick substitution, say $\sin\theta=u$. Then our equation becomes
And we've solved for $u$! But we werent solving for $u$, we're supposed to be solving for $\theta$. So we undo our substitution to finish up.
$$\sin\theta=\frac{1}{2}\implies \theta = \frac{\pi}{6},\frac{5\pi}{6}$$ $$\sin\theta=1\implies \theta = \frac{\pi}{2}$$
Plus any multiple of $2\pi$, of course. So the general solution would be:
$$ \theta = \frac{\pi}{6} + 2\pi k \text{ or } \frac{5\pi}{6} + 2\pi k \text{ or } \frac{\pi}{2} $$
for any $k\in \mathbb{Z}$
Note that you don't have to factor here, it is a preference. You could have used the quadratic formula or any other method you've learned to solve quadratics.
Example: Using the quadrtic formula
Find all solutions to $2\cos^2\theta + \cos\theta = 1$ on the interval $0\leq \theta < 2\pi$
::solution First, make the substitution $\cos\theta = u$. This yields $2u^2+u=1$ leading us to $2u^2+u-1=0$. Identifying $a=2, b=1, c=-1$ we have
Undoing the substitution we have
$$\cos\theta = \frac{1}{2} \implies \theta = \frac{\pi}{3},\frac{5\pi}{3}$$ $$\cos\theta = -1 \implies \theta = \pi$$
Adding $2\pi$ to any of these would go outside of our interval $0\leq \theta < 2\pi$. Therefore our solution set is
$$\{ \frac{\pi}{3},\frac{5\pi}{3},\pi\}$$ :::
MyOpenMath: Careful, you won't always find a solution
Using Pythagorean Identities
It can also happen that we run into an equation that is not only resistant to factoring, but isn't even using the same trig functions.
Example: In need of Pythagoras
Find the general solution to $2\cos^2\theta + \sin\theta - 1 = 0$.
Show Solution
A good rule of thumb is: if you can turn the equation into all of one type of trig function, you should. Here, the mixing of sine and cosine presents a problem in that we can't just substitute and factor. But there are several relationships and identities we've learned and can use here. One that seems the most relevant to me would be the OG $\sin^2\theta + \cos^2\theta = 1$. If we shuffle it around a bit we can get $\cos^2\theta = 1 - \sin^2\theta$ and that can be substituted into our equation.
Now we have an equation that only has sines in it. We can proceed with our normal methods.
for any $k\in \mathbb{Z}$.
Example: Or was it Hippasus
Solve the equation $2\sin^2\theta = 1 - \cos\theta$ on the interval $0\leq \theta < 2\pi$
::solution Begin by using $\sin^2\theta + \cos^2\theta = 1$.
Now this is a quadratic in cosine. Let $u=\cos\theta$.
So $u=1$ or $u=-\frac{1}{2}$, which means
$$ \cos\theta=1 \quad\text{or}\quad \cos\theta=-\frac{1}{2} $$
On $0\leq\theta<2\pi$:
- $\cos\theta=1$ at $\theta=0$
- $\cos\theta=-\frac{1}{2}$ at $\theta=\frac{2\pi}{3},\frac{4\pi}{3}$
So the solution set is
$$ \{0,\frac{2\pi}{3},\frac{4\pi}{3}\} $$ :::