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2cos2x=-sinx solve for x over
f(x) = x² + 2Cos²x, find f ' (x)
a) 2(x+cos x) b) x - sin x c) 2x + sin x d)2(x - sin2x) I got neither of these answers, since
3 answers
asked by
Terry
1,103 views
2cos2x=-sinx
solve for x over the interval [-pi,+pi] x= 0.6349radians x=2.507radians x= 1.003 radians x= 2.138 radians
10 answers
asked by
SkiMasktheSlumpGod
743 views
Prove the following:
[1+sinx]/[1+cscx]=tanx/secx =[1+sinx]/[1+1/sinx] =[1+sinx]/[(sinx+1)/sinx] =[1+sinx]*[sinx/(sinx+1)]
3 answers
asked by
Anonymous
996 views
Solve the equation of the interval (0, 2pi)
cosx=sinx I squared both sides to get :cos²x=sin²x Then using tri indentites I came
4 answers
asked by
Brian
851 views
solve each equation for 0=/<x=/<2pi
sin^2x + 5sinx + 6 = 0? how do i factor this and solve? 2sin^2 + sinx = 0 (2sinx - 3)(sinx
7 answers
asked by
sh
1,011 views
the problem is
2cos^2x + sinx-1=0 the directions are to "use an identity to solve each equation on the interval [0,2pi). This is
3 answers
asked by
alex
917 views
Mathematics - Trigonometric Identities - Reiny, Friday, November 9, 2007 at 10:30pm
(sinx - 1 -cos^2x) (sinx + 1 - cos^2x) should
3 answers
asked by
Anonymous
823 views
Solving Trigonometic Equations
solve for x and give the answers as a equations : ( by radian) 1)cos(sinx)=1 <<<and thanks >>> We
0 answers
asked by
abdo
718 views
Simplify sin x cos^2x-sinx
Here's my book's explanation which I don't totally follow sin x cos^2x-sinx=sinx(cos^2x-1)
1 answer
asked by
Tara
1,095 views
solve to four decimals places
5.0118 sinx - 3.1105=0 :for all real x values I will be happy to critique your work. Solve for
2 answers
asked by
Kate
576 views