Solve for t arcsin(t^2-2t)=pi/2

$\arcsin (t^{2} - 2 t) = \frac{π}{2}$

Step 1

Take the inverse arcsine of both sides of the equation to extract $t$ from inside the arcsine.

$t^{2} - 2 t = \sin (\frac{π}{2})$

Step 2

Simplify the right side.

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$t^{2} - 2 t = 1$

Step 3

Subtract $1$ from both sides of the equation.

$t^{2} - 2 t - 1 = 0$

Step 4

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 5

Substitute the values $a = 1$ , $b = - 2$ , and $c = - 1$ into the quadratic formula and solve for $t$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (1 \cdot - 1)}}{2 \cdot 1}$

Step 6

Simplify.

Simplify the numerator.

Raise $- 2$ to the power of $2$ .

$t = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 1}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 1$ .

Multiply $- 4$ by $1$ .

$t = \frac{2 \pm \sqrt{4 - 4 \cdot - 1}}{2 \cdot 1}$

Multiply $- 4$ by $- 1$ .

$t = \frac{2 \pm \sqrt{4 + 4}}{2 \cdot 1}$

Add $4$ and $4$ .

$t = \frac{2 \pm \sqrt{8}}{2 \cdot 1}$

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

$t = \frac{2 \pm \sqrt{4 (2)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

$t = \frac{2 \pm \sqrt{2^{2} \cdot 2}}{2 \cdot 1}$

Pull terms out from under the radical.

$t = \frac{2 \pm 2 \sqrt{2}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$t = \frac{2 \pm 2 \sqrt{2}}{2}$

Simplify $\frac{2 \pm 2 \sqrt{2}}{2}$ .

$t = 1 \pm \sqrt{2}$

Step 7

The final answer is the combination of both solutions.

$t = 1 + \sqrt{2}, 1 - \sqrt{2}$

Step 8

Exclude the solutions that do not make $\arcsin (t^{2} - 2 t) = \frac{π}{2}$ true.

$t = 1 + \sqrt{2}$

Step 9

The result can be shown in multiple forms.

Exact Form:

$t = 1 + \sqrt{2}$

Decimal Form:

$t = 2.41421356 \dots$

Solve for t arcsin(t^2-2t)=pi/2

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