Brannslukker
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[h=2]tan(x)*sin(x)=tan(x)=cos(x), help solve equation![/h]Its an equation i can not figure out, please help. x is defined from 0 to 2pi radians
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tan(x)*sin(x)=tan(x)=cos(x), help solve equation!
- Thread starter Brannslukker
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HallsofIvy
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Do you at least know that \(\displaystyle tan(x)= \frac{sin(x)}{cos(x)}\)? So that your equations are \(\displaystyle \frac{sin(x)}{cos(x)}sin(x)= \frac{sin(x)}{cos(x)}= cos(x)\).
Multiplying each part by cos(x), \(\displaystyle sin^2(x)= sin(x)= cos^2(x)\) which is, of course, the same as \(\displaystyle sin^2(x)= sin(x)= 1- sin^2(x)\).
From \(\displaystyle sin^2(x)= 1- sin^2(x)\) so that \(\displaystyle 2 sin^2(x)= 1\), \(\displaystyle sin(x)= \frac{\sqrt{2}}{2}\).
From \(\displaystyle sin^2(x)= sin(x)\), either \(\displaystyle sin(x)= 0\) or \(\displaystyle sin(x)= 1\).
Hopefully, you realize that there are no values of x that satisfy both of those conditions.
Multiplying each part by cos(x), \(\displaystyle sin^2(x)= sin(x)= cos^2(x)\) which is, of course, the same as \(\displaystyle sin^2(x)= sin(x)= 1- sin^2(x)\).
From \(\displaystyle sin^2(x)= 1- sin^2(x)\) so that \(\displaystyle 2 sin^2(x)= 1\), \(\displaystyle sin(x)= \frac{\sqrt{2}}{2}\).
From \(\displaystyle sin^2(x)= sin(x)\), either \(\displaystyle sin(x)= 0\) or \(\displaystyle sin(x)= 1\).
Hopefully, you realize that there are no values of x that satisfy both of those conditions.
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tan(x)*sin(x)=tan(x)=cos(x), help solve equation!
Its an equation i can not figure out, please help. x is defined from 0 to 2pi radians
That's not AN equation there - you have two equations.
Moreover, what do you want to do with those equations? Prove identities? Solve for 'x'?