Solving Trigonometric Equations
Trigonometric Equations are equations that involve trigonometric ratios such as \(\sin(x)\) , \(\cos(x)\) and \(\tan (x)\) in place of standard variables such as \(x\). Trigonometric equations can either be solved algebraically by hand or graphically with technology.
There are often multiple solutions for a trigonometric equation, given that different angles for a common trigonometric ratio can carry the same solution. For example:
\(\sin\left(\cfrac{3\pi}{4}\right) = \cfrac{\sqrt{2}}{2}\)
As shown above, different angles for the same trigonometric function (cosine) carries the same solution \(\left(\cfrac{\sqrt{2}}{2}\right)\). This occurs because trigonometric functions are periodic. Ensure that you find all solutions that lie in the domain of interest.
We can refer to the Unit Circle/CAST diagram to determine the signs of each trigonometric ratio in relation to what quadrant its located:
Although an equation may have one variable like \(\theta\), each trig ratio in an equation (like \(\sin\theta\), \(\cos\theta\)) can be through of as separate terms because you cannot combine them as like terms. Therefore, we sometimes need to convert between different trig ratios until there is only one ratio.
We can use the list of trigonometric identities in order to convert certain ratios and make it possible to determine the solutions to an equation. However, terms like \(\sin(x)\) and \(\sin^2(x)\) are similar terms like \(x\) and \(x^2\). We can use techniques like factoring to solve these equations!
We can review Trigonometric Identities in order to help us solve Trigonometric Equations.
Example
Determine approximate solutions for the equation, \(\cos(x) - \cfrac{1}{2} = 0\),on the interval \(x\in[0,2\pi]\), to the nearest hundredth of a radian.
First, we can place the terms on opposite sides to isolate for the term with the variable:
We can determine that this ratio lies in both Quadrants 1 and 4 since those are the quadrants where the sine function is positive.
In order to get our first solution, we can then set our calculators to radians and calculate the inverse of sine on both sides:
\(x_1 = \cos\left(\cfrac{1}{6}\right)^{-1}\)
\(x_1 = 1.40 \; [\text{rads}]\)
Since the second solution is located in Quadrant 4, we can determine its value by subtracting the first solution from \(2\pi\):
\(x_2 = 2\pi [\text{rads}]\; [\text{rads}] - 1.40 \; [\text{rads}]\)
\(x_2 = 4.88 \; [\text{rads}]\)
Therefore, we can determine our \(2\) solutions are \(\boldsymbol{1.40\; [\textbf{rads}]}\) and \(\boldsymbol{4.88 \; [\textbf{rads}]}\).
\(\tan(2x) + \sqrt{3} = 0\)
First, we can let \(2x = \theta\). We can then rewrite the equation as such:
Next, we can place the terms on opposite sides to isolate for the term with the variable:
We can determine that this ratio lies in both Quadrants 2 and 4 since those are the quadrants where the tangent function is negative.
In order to get our first solution, we can then set our calculators to radians and calculate the inverse of sine on both sides:
\(\theta_1 = \tan(-\sqrt{3})^{-1}\)
\(\theta_1 = -1.047197551 \; [\text{rads}]\)
We can now substitute \(2x\) for \(\theta\) and calculate \(\theta_1\):
\(2x_1 = -1.047197551 \; [\text{rads}]\)
\(x_1 = -0.52 \; [\text{rads}]\)
Since one of the solutions is located in Quadrant \(4\), we can convert this solution to a positive value by adding it to \(2π\):
\(x_1 = -0.52 \; [\text{rads}] + 2\pi \; [\text{rads}]\)
\(x_1 = 5.76 \; [\text{rads}]\)
Since the second solution is located in Quadrant \(2\), we can determine its value by subtracting \(\pi\) from the first solution:
\(x_2 = 5.76 \; [\text{rads}] - \pi\; [\text{rads}]\)
\(x_2 = 2.62 \; [\text{rads}]\)
Therefore, we can determine our \(2\) solutions are \(\boldsymbol{5.76 \; [\textbf{rads}}]\) and \(\boldsymbol{2.16 \; [\textbf{rads}]}\).
\(\csc(x) + \cos(x) + 1 = \cot(x)\sin(x)\)
First, we can use trigonometric identities to convert \(\csc(x)\) and \(\cot(x)\):
\(\cot(x) = \cfrac{\cos(x)}{\sin(x)}\)
\(\cfrac{1}{\sin(x)} + \cos(x) + 1 = \cfrac{\cos(x)}{\sin(x)} \cdot \sin(x)\)
We can now simplify the equation on the right side:
\(\cfrac{1}{\sin(x)} + \cos(x) + 1 = \cos(x)\)
We can simplify the expression further by \(\cos(x) + 1\) from the left side to the right side and cancelling terms:
\(\cfrac{1}{\sin(x)} = \cancel{\cos(x)} \cancel{-\cos(x)} -1\)
\(\cfrac{1}{\sin(x)} = -1\)
Next, we can multiply both sides by \(-\sin(x)\):
\(\left(\cfrac{1}{\cancel{\sin(x)}}\right)(-\cancel{\sin(x)}) = (\cancel{-}1)(\cancel{-}\sin(x))\)
\(\sin(x) = -1\)
Then, we can calculate the inverse of sine on both sides:
\(x = \sin(-1)^{-1}\)
\(x = -1.57\;[\text{rads}]\)
We can convert this solution to a positive value by adding it to \(2π\):
\(x = 2\pi\; [\text{rads}] - 1.57\; [\text{rads}]\)
\(x = 4.71\; [\text{rads}]\)
Therefore, we can determine our solution is \(\boldsymbol{4.71 \; [\textbf{rads}]}\), located at the negative \(y\)-axis.