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amy0799
 one year ago
Let f(x) = sinxcosx + 0 for
amy0799
 one year ago
Let f(x) = sinxcosx + 0 for

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amy0799
 one year ago
Best ResponseYou've already chosen the best response.0\[0 \le x \le \frac { 3\pi }{ 2} \] find all the values for which f'(x)=1

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1the question really contains + 0....?

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0oops its suppose to be sinxcosx+x

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1ok.... so you need the product rule to do the sin(x)cos(x) or you can use a double angle substitution... do you know sin(2x) = 2 sin(x)cos(x)..?

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0ill do the product rule f'(x)=cosxcosx+cosx(sinx)+1

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1well not quite u = sin(x) v = cos(x) u' = cos(x) v' = sin(x) so the derivative is \[\frac{dy}{dx} = \sin(x) \times  \sin(x) + \cos(x) \times \cos(x) + 1\] or \[\frac{dy}{dx} = \cos^2(x)  \sin^2(x) + 1\] does that make sense

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0ooh I see what I did wrong

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1so the derivative is equal to 1 then \[1 = \cos^2(x)  \sin^2(x) + 1~~~or~~~0 = \cos^2(x)  \sin^2(x)\] I'd now make a substitution of \[\sin^2(x) = 1  \cos^2(x)\] so you have \[0 = \cos^2(x) (1  \cos^2(x) \] or \[0 = 2\cos^2(x)  1\] now you should be able to solve for x

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0what do I put in x? 3pi/2? @campbell_st

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1no you solve so \[1 = 2\cos^2(x)\] then \[\frac{1}{2} = \cos^2(x)\] so \[\pm \frac{1}{\sqrt{2}} = \cos(x)\] this is an exact value.... that is equal to \[x = \frac{\pi}{4}\] in the 1st quadrant so there are angles in the 2nd quadrant \[\pi  \frac{\pi}{4}\] and 3rd quadrant \[\pi + \frac{\pi}{4}\] just simplify the angles in the 2nd and 3rd quadrants by writing them as improper fractions.

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0how did you get pi/4? @campbell_st

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1because I know \[\frac{1}{\sqrt{2}} = \frac{\pi}{4}\] its called an exact value, you may know it with a rationalized denominator as \[\frac{\sqrt{2}}{2} = \frac{\pi}{4}\]

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1but if you want to check, use a calculator and just enter \[\cos^{1}(\frac{1}{\sqrt{2}})\]

amy0799
 one year ago
Best ResponseYou've already chosen the best response.0ooh ok so what do you mena by simplify the angles in the 2nd and 3rd quadrants by writing them as improper fractions?

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1well the angles are normally written using fractions so 3rd quadrant \[\pi + \frac{\pi}{4} = \frac{5\pi}{4}\] you need to work out the 2nd quadrant angle
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