# Solving for X with sin²

• Dec 22nd 2012, 05:19 AM
acwilson96
Solving for X with sin²
So I missed a week of school and the holidays started and to catch up I have been given some work and I was struggling to complete this as I haven't had any tuition on how to:

sin²X=0.25

Its been bugging me and I don't know what to do to get rid of the sin². Had it just been sin I would have 'inverse sined' a quarter

All help appreciated
• Dec 22nd 2012, 05:26 AM
Plato
Re: Solving for X with sin²
Quote:

Originally Posted by acwilson96
So I missed a week of school and the holidays started and to catch up I have been given some work and I was struggling to complete this as I haven't had any tuition on how to:
sin²X=0.25
Its been bugging me and I don't know what to do to get rid of the sin². Had it just been sin I would have 'inverse sined' a quarter

Surely you realize that $\sin(x)=\pm 0.5~?$
• Dec 22nd 2012, 05:33 AM
acwilson96
Re: Solving for X with sin²
No sorry i didnt, im doing higher maths(SQA) and im not sure that is in the syllabus or perhaps i just missed it. But what steps do I do to get rid of sin²?
• Dec 22nd 2012, 06:01 AM
Plato
Re: Solving for X with sin²
Quote:

Originally Posted by acwilson96
No sorry i didnt, im doing higher maths(SQA) and im not sure that is in the syllabus or perhaps i just missed it. But what steps do I do to get rid of sin²?

If $u^2=0.25$ then $u=\pm 0.5~.$
• Dec 22nd 2012, 06:05 AM
acwilson96
Re: Solving for X with sin²
oh right yes, i understand that but i thought if you square root then it would be sin(root x )= 0.25
• Dec 22nd 2012, 06:18 AM
Plato
Re: Solving for X with sin²
Quote:

Originally Posted by acwilson96
oh right yes, i understand that but i thought if you square root then it would be sin(root x )= 0.25

Well $\sin^2(x)=[\sin(x)]^2$.

Is the question really $\sin(\sqrt{x})=0.25~?$.

Which is it?
• Dec 22nd 2012, 06:31 AM
acwilson96
Re: Solving for X with sin²
Ok i guess its just a rule you need to know that sin²x=(sinx)²
Il keep that in mind. So the answer would inverse sin of 1/2
• Dec 22nd 2012, 06:37 AM
Plato
Re: Solving for X with sin²
Quote:

Originally Posted by acwilson96
Ok i guess its just a rule you need to know that sin²x=(sinx)²
Il keep that in mind. So the answer would inverse sin of 1/2

$x=\pm\arcsin(0.5)$