Solve Equations Including Inverse Trigonometric Functions
Learn how to solve equations that involve inverse trigonometric functions, such as arcsin, arccos, and arctan.
This guide is designed for Grade 12 math students and includes step-by-step solutions to common inverse trig equations. Each solution is also verified using a graphical method by plotting both sides of the equation on
the same coordinate system. The point of intersection gives an approximate solution, helping students visualize the equation and understand the concept more deeply.
Question 1
Solve for x the equation \( 3 \arcsin(x) = \dfrac{\pi}{2} \).
Solution
Divide both sides of the equation by 3.
\[
\arcsin(x) = \dfrac{\pi}{6}
\]
Apply sin to both sides and simplify.
\[
\sin(\arcsin(x)) = \sin\left(\dfrac{\pi}{6}\right)
\]
The above simplify to
\[
x = \dfrac{1}{2}
\]
Because of the domain of arcsin(x), we need to verify that the solution obtained is valid.
Check solution
\[
x = \dfrac{1}{2}
\]
Right side of equation: \( 3 \arcsin\left(\dfrac{1}{2}\right) = 3 \cdot \dfrac{\pi}{6} = \dfrac{\pi}{2} \)
Left side of equation: \( \dfrac{\pi}{2} \)
The solution to the above equation is \( x = \dfrac{1}{2} \)
The graphical approximation to the solution to the given equation is shown below. The x coordinate of the point of intersection of the graphs made up the left side and the right of the given equation is 0.5 which is the solution found analytically.
Question 2
Solve for x the equation \( 3 \cot(\arccos(x)) = 2 \).
Solution
Divide both sides of the given equation by 3 and simplify.
\[
\cot(\arccos(x)) = \dfrac{2}{3}
\]
Let \( A = \arccos(x) \)
and apply cos to both sides to obtain.
\[
\cos(A) = \cos(\arccos(x)) = x
\]
Using definition of A above, the equation may be written as
\[
\cot(A) = \dfrac{2}{3}
\]
Use \( \cot(A) = \dfrac{2}{3} \) to construct a right triangle and find \( \cos(A) \). Find hypotenuse h first.
\[
h = \sqrt{13}
\]
We now use the same triangle shown above to find \( \cos(A) \).
\[
x = \cos(A) = \dfrac{2}{\sqrt{13}} \approx 0.55
\]
The graphical approximation to the solution to the given equation is shown below.
Question 3
Solve for x the equation \( \arcsin(x) = \arccos(x) \).
Solution
Apply sin function to both sides.
\[
\sin(\arcsin(x)) = \sin(\arccos(x))
\]
Simplify left side using the identity \( \sin(\arcsin(A)) = A \).
\[
x = \sin(\arccos(x))
\]
Let \( A = \arccos(x) \), then \( \cos(A) = x \)
\[
\sin(\arccos(x)) = \sin(A) = \pm \sqrt{1 - x^2}
\]
Use the above to rewrite the given equation in algebraic form.
\[
x = \pm \sqrt{1 - x^2}
\]
Square both sides.
\[
x^2 = 1 - x^2
\]
\[
2x^2 = 1
\]
\[
x = \pm \dfrac{1}{\sqrt{2}}
\]
Because of the domain of the arccos function and we also squared both sides of the equation, we need to verify the solutions and eliminate any invalid (extraneous) ones.
Check solution
1) \( x = \dfrac{1}{\sqrt{2}} \)
Left side: \( \arcsin\left(\dfrac{1}{\sqrt{2}}\right) = \dfrac{\pi}{4} \)
Right side: \( \arccos\left(\dfrac{1}{\sqrt{2}}\right) = \dfrac{\pi}{4} \)
\( x = \dfrac{1}{\sqrt{2}} \) is a solution to the given equation.
2) \( x = -\dfrac{1}{\sqrt{2}} \)
Left side: \( \arcsin\left(-\dfrac{1}{\sqrt{2}}\right) = -\dfrac{\pi}{4} \)
Right side: \( \arccos\left(-\dfrac{1}{\sqrt{2}}\right) = \dfrac{3\pi}{4} \)
\( x = -\dfrac{1}{\sqrt{2}} \) is not a solution to the given equation.
The graphical approximation to the solution to the given equation is shown below. The x coordinate of the point of intersection 0.71 is close to \( \dfrac{1}{\sqrt{2}} \).
Question 4
Solve for x the equation \( \arccos(x) = \arcsin(x) + \dfrac{\pi}{2} \)
Solution
Apply cos function to both sides.
\[
\cos(\arccos(x)) = \cos\left(\arcsin(x) + \dfrac{\pi}{2}\right)
\]
Simplify left side using the identity \( \cos(\arccos(A)) = A \).
\[
x = \cos\left(\arcsin(x) + \dfrac{\pi}{2}\right)
\]
Expand the right side using the identity \( \cos(a + b) = \cos(a)\cos(b) - \sin(a)\sin(b) \).
\[
x = \cos(\arcsin(x))\cos\left(\dfrac{\pi}{2}\right) - \sin(\arcsin(x))\sin\left(\dfrac{\pi}{2}\right)
\]
Use \( \cos\left(\dfrac{\pi}{2}\right) = 0 \), \( \sin(\arcsin(x)) = x \), and \( \sin\left(\dfrac{\pi}{2}\right) = 1 \)
\[
x = -x
\]
\[
2x = 0
\]
\[
x = 0
\]
Check solution
Left side: \( \arccos(0) = \dfrac{\pi}{2} \)
Right side: \( \arcsin(0) + \dfrac{\pi}{2} = \dfrac{\pi}{2} \)
\( x = 0 \) is a solution to the given equation
The graphical approximation to the solution to the given equation is shown below. The x coordinate of the point of intersection is equal to 0 exactly as the value calculated analytically above.
Question 5
Solve for x the equation \( \arccos(2x) = \dfrac{\pi}{3} + \arccos(x) \)
Solution
Apply cosine function to both sides.
\[
\cos(\arccos(2x)) = \cos\left(\dfrac{\pi}{3} + \arccos(x)\right)
\]
Simplify left side using identity and expand right side.
Left side: \( \cos(\arccos(2x)) = 2x \)
Right side:
\[
\cos\left(\dfrac{\pi}{3} + \arccos(x)\right) = \cos\left(\dfrac{\pi}{3}\right)\cos(\arccos(x)) - \sin\left(\dfrac{\pi}{3}\right)\sin(\arccos(x))
\]
\[
= \dfrac{1}{2}x \pm \dfrac{\sqrt{3}}{2}\sqrt{1 - x^2}
\]
Rewrite the equation with radical on the right side.
\[
2x = \dfrac{x}{2} \pm \dfrac{\sqrt{3}}{2}\sqrt{1 - x^2}
\]
\[
\dfrac{3x}{2} = \pm \dfrac{\sqrt{3}}{2} \sqrt{1 - x^2}
\]
Square both sides.
\[
\dfrac{9x^2}{4} = \dfrac{3}{4}(1 - x^2)
\]
\[
\dfrac{12x^2}{4} = \dfrac{3}{4}
\]
\[
x^2 = \dfrac{1}{4}
\]
\[
x = \pm \dfrac{1}{2}
\]
Check solution
1) \( x = \dfrac{1}{2} \)
Left side: \( \arccos(2x) = \arccos(1) = 0 \)
Right side: \( \dfrac{\pi}{3} + \arccos\left(\dfrac{1}{2}\right) = \dfrac{\pi}{3} + \dfrac{\pi}{3} = \dfrac{2\pi}{3} \)
\( x = \dfrac{1}{2} \) is NOT a solution
2) \( x = -\dfrac{1}{2} \)
Left side: \( \arccos(-1) = \pi \)
Right side: \( \dfrac{\pi}{3} + \arccos\left(-\dfrac{1}{2}\right) = \dfrac{\pi}{3} + \dfrac{2\pi}{3} = \pi \)
\(x = -1/2 \) is a solution to the given equation.
The graphical approximation to the solution to the given equation is shown below. The point of intersection has an x coordinate equal to -0.5 which is exactly the solution found analytically.
More References and links