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**fusilli_jerry89****Member**- Registered: 2006-06-23
- Posts: 86

does anyone know how to get this question? I got pi/2 and 3pi/3.

√2 sin² -sin = 0 Solve for values of sin. 0<x<2pi

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,562

How about 0.707

**igloo** **myrtilles** **fourmis**

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Forget the sins for the moment and solve it as if it was a simple quadratic equation.

√2 sin²x -sinx = 0

sinx(√2 sinx -1) = 0

sinx = 0 or (√2 sinx -1) = 0

If (√2 sinx -1) = 0, then √2 sinx = 1 and so sinx = 1/√2.

So now we have sinx = 0 or 1/√2.

Therefore, x = 0, π/4, 3π/4, π, and 2π.

0 and 2π may not be answers, depending on whether the limits of the range are included.

Why did the vector cross the road?

It wanted to be normal.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,562

Right, because .707 is 1/√2.

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,562

Also half of the square root of 2 is the same as 1 over the square root of 2.

**igloo** **myrtilles** **fourmis**

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