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 $\displaystyle \sin(x)=\pm 0.5~?$

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?

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 $\displaystyle u^2=0.25$ then $\displaystyle u=\pm 0.5~.$

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

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

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

Which is it?

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

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

$\displaystyle x=\pm\arcsin(0.5)$