# Equation.Solve the equation square root 2*sin x - 1 = 0, 0=<x=<pi.

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### 2 Answers

You need to isolate the trigonometric function to one side, such that:

`sqrt2*sin x = 1 => sin x = 1/(sqrt 2) => sin x = sqrt2/2`

You need to solve the equation for `x in (0,pi)` , hence, you will need to find two solutions for x, such that:

`x = pi/4 if x in (0,pi/2)`

`x = pi/2 + pi/4 => x = (3pi)/4 if x in (pi/2,pi)`

**Hence, evaluating the solutions to the given equation, under the given conditions, yields that `sin x = sqrt2/2` at **`x = {pi/4, (3pi)/4}.`

This is an elementary trigonometric equation.

We'll shift 1 to the right:

sqrt2*sin x = 1

sin x = 1/sqrt2

x = arcsin (1/sqrt2)

Since the sine function is positive over the range [0,pi], we'll get

x = pi/4 or x = pi-pi*4 = 3pi/4