Solve Equations Including Inverse Trigonometric Functions

Solve for $x$ the equation $3 \arcsin(x) = \dfrac{\pi}{2}$.

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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 simplifies to

$$ x = \dfrac{1}{2} $$

Because of the domain of $\arcsin(x)$, we need to verify that the solution obtained is valid.

Check solution:

For $x = \dfrac{1}{2}$

• 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 making up the left side and the right side of the given equation is $0.5$, which is the solution found analytically.

Solve for $x$ the equation $3 \cot(\arccos(x)) = 2$.

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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 the 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 the hypotenuse $h$ first.

$$ h = \sqrt{2^2 + 3^2} = \sqrt{13} $$

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.

Solve for $x$ the equation $\arcsin(x) = \arccos(x)$.

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Solution:

Apply the $\sin$ function to both sides.

$$ \sin(\arcsin(x)) = \sin(\arccos(x)) $$

Simplify the left side using the identity $\sin(\arcsin(A)) = A$.

$$ x = \sin(\arccos(x)) $$

Let $A = \arccos(x)$, then $\cos(A) = x$. Using right triangle trigonometry (or the Pythagorean identity):

$$ \sin(\arccos(x)) = \sin(A) = \pm \sqrt{1 - x^2} $$

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 the fact that we squared both sides of the equation, we need to verify the solutions and eliminate any invalid (extraneous) ones.

Check solution:

1) For $x = \dfrac{1}{\sqrt{2}}$

• Left side: $\arcsin\left(\dfrac{1}{\sqrt{2}}\right) = \dfrac{\pi}{4}$
• Right side: $\arccos\left(\dfrac{1}{\sqrt{2}}\right) = \dfrac{\pi}{4}$

$\implies x = \dfrac{1}{\sqrt{2}}$ is a solution to the given equation.

2) For $x = -\dfrac{1}{\sqrt{2}}$

• Left side: $\arcsin\left(-\dfrac{1}{\sqrt{2}}\right) = -\dfrac{\pi}{4}$
• Right side: $\arccos\left(-\dfrac{1}{\sqrt{2}}\right) = \dfrac{3\pi}{4}$

$\implies x = -\dfrac{1}{\sqrt{2}}$ 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 $\dfrac{1}{\sqrt{2}}$.

Solve for $x$ the equation $\arccos(x) = \arcsin(x) + \dfrac{\pi}{2}$.

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Solution:

Apply the $\cos$ function to both sides.

$$ \cos(\arccos(x)) = \cos\left(\arcsin(x) + \dfrac{\pi}{2}\right) $$

Simplify the left side using the identity $\cos(\arccos(A)) = A$.

$$ x = \cos\left(\arcsin(x) + \dfrac{\pi}{2}\right) $$

Expand the right side using the compound angle 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 the known values: $\cos\left(\dfrac{\pi}{2}\right) = 0$, $\sin(\arcsin(x)) = x$, and $\sin\left(\dfrac{\pi}{2}\right) = 1$.

$$ x = \cos(\arcsin(x)) \cdot 0 - x \cdot 1 $$

$$ x = -x $$

$$ 2x = 0 $$

$$ x = 0 $$

Check solution:

• Left side: $\arccos(0) = \dfrac{\pi}{2}$
• Right side: $\arcsin(0) + \dfrac{\pi}{2} = 0 + \dfrac{\pi}{2} = \dfrac{\pi}{2}$

$x = 0$ is a valid 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.

Solve for $x$ the equation $\arccos(2x) = \dfrac{\pi}{3} + \arccos(x)$.

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Solution:

Apply the cosine function to both sides.

$$ \cos(\arccos(2x)) = \cos\left(\dfrac{\pi}{3} + \arccos(x)\right) $$

Simplify the left side using the inverse identity, and expand the right side using the compound angle formula.

Left side: $\cos(\arccos(2x)) = 2x$

Right side:

$$ \cos\left(\dfrac{\pi}{3} + \arccos(x)\right) = \cos\left(\dfrac{\pi}{3}\right)\cos(\arccos(x)) - \sin\left(\dfrac{\pi}{3}\right)\sin(\arccos(x)) $$

$$ = \dfrac{1}{2}x \pm \dfrac{\sqrt{3}}{2}\sqrt{1 - x^2} $$

Equating both sides, rewrite the equation with the radical isolated 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.

$$ \dfrac{9x^2}{4} = \dfrac{3}{4}(1 - x^2) $$

Multiply by 4 to clear denominators:

$$ 9x^2 = 3(1 - x^2) $$

$$ 9x^2 = 3 - 3x^2 $$

$$ 12x^2 = 3 $$

$$ x^2 = \dfrac{1}{4} $$

$$ x = \pm \dfrac{1}{2} $$

Check solution:

1) For $x = \dfrac{1}{2}$

• Left side: $\arccos\left(2 \cdot \dfrac{1}{2}\right) = \arccos(1) = 0$
• Right side: $\dfrac{\pi}{3} + \arccos\left(\dfrac{1}{2}\right) = \dfrac{\pi}{3} + \dfrac{\pi}{3} = \dfrac{2\pi}{3}$

$\implies x = \dfrac{1}{2}$ is NOT a solution.

2) For $x = -\dfrac{1}{2}$

• Left side: $\arccos\left(2 \cdot -\dfrac{1}{2}\right) = \arccos(-1) = \pi$
• Right side: $\dfrac{\pi}{3} + \arccos\left(-\dfrac{1}{2}\right) = \dfrac{\pi}{3} + \dfrac{2\pi}{3} = \pi$

$\implies x = -\dfrac{1}{2}$ is a valid 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.