could this be a typo? in a question sheet? 2cosecx-1=0???

Ah ok poobah, yeah i realized its csc right after i posted the first time

 

[math]2csc(x) - 1 = 0[/math]

[math]{\frac{1}{sin(x)}} = 0.5[/math]

[math]sin^{-1}(x) = 2[/math]

[math]x = sin(2)[/math]

[math]x = no solution[/math]

 

as you can see then there are no solutions (since sin(x) is 1 at max.)

Edited March 29, 200916 yr by max.yevs

No - written as....... 2csc-1=0

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Merged post follows:

Consecutive posts merged

as i thought - unsolveable - prob just a typo - but ty

[math]sin^{-1}(x) = 2[/math]

[math]x = sin(2)[/math]

[math]x = no solution[/math]

 

as you can see then there are no solutions (since sin(x) is 1 at max.)

sin(2) definitely has a value, approximately 0.9.

Ah ok poobah, yeah i realized its csc right after i posted the first time

 

[math]2csc(x) - 1 = 0[/math]

[math]{\frac{1}{sin(x)}} = 0.5[/math]

[math]sin^{-1}(x) = 2[/math]

[math]x = sin(2)[/math]

[math]x = no solution[/math]

 

as you can see then there are no solutions (since sin(x) is 1 at max.)

 

This is the step you got wrong...

[math]sin^{-1}(x) = 2[/math]

as tree pointed out, there is an x that solves this step -- you took the inverse of the sine function too early...

 

 

[math]2\csc x -1 =0[/math]

[math]2\frac{1}{\sin x} = 1 [/math]

[math]\frac{1}{\sin x} = \frac{1}{2}[/math]

[math]\sin x = 2[/math]

 

and now you conclude that x has no solution because sin is limited to values between -1 and 1.

ah right...

 

that's what happens when you go from step1 to step 2 in your head...