Video: Solving Equations Involving an Inverse Trigonometric Function

Solve sin⁻¹ π‘₯ + πœ‹ = 3πœ‹/2.

01:33

Video Transcript

Solve sin inverse of π‘₯ plus πœ‹ equals three πœ‹ over two.

To solve, we need to isolate this π‘₯-value. The first thing we would do is subtract πœ‹ from both sides of the equation. On the left, πœ‹ minus πœ‹ cancels out. And we’re left with the sin inverse of π‘₯. On the right, we have three πœ‹ over two minus πœ‹. We can rewrite πœ‹ as two πœ‹ over two.

And then we have something that says three πœ‹ over two minus two πœ‹ over two. Three πœ‹ minus two πœ‹ equals πœ‹. And we keep the denominator of two. And so, the sin inverse of π‘₯ equals πœ‹ over two. To get rid of the sin inverse, we need to take the sin of the sin inverse of π‘₯. And if we take the sin of the left side of the equation, we’ll need to take the sin of the right side of the equation.

The sin of the sin inverse of π‘₯ equals π‘₯. And that means π‘₯ equals the sin of πœ‹ over two. We know that the sin of πœ‹ over two equals one. Our π‘₯ must equal one.

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