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#1 2006-06-23 12:00:11

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

Trig

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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#2 2006-08-31 21:31:10

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

Re: Trig

How about 0.707


igloo myrtilles fourmis

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#3 2006-08-31 23:12:49

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

Re: Trig

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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#4 2006-09-01 01:58:49

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

Re: Trig

Right, because .707 is 1/√2.


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#5 2006-09-01 02:01:19

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

Re: Trig

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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