Trig and differentiate Q (1 Viewer)

sdd10 · Apr 3, 2009

I need help with the following 2 Questoins

Solve cos 2x = 2 + cox x for 0<x<180

Diffferentiate y = ln[(x+1)^3/sqr rt (x-1)]

Thanks in advance

shaon0 · Apr 3, 2009

Can't understand the one above it.
1) Diffferentiate y = ln[(x+1)^3/sqr rt (x-1)]
y=ln((x+1)^3)-ln((x-1)^(1/2))
= 3ln(x+1)-0.5ln(x-1)
=3/(x+1)-1/(2ln(x-1))

Drongoski · Apr 3, 2009

$cos\ 2x\ =\ 2 \ +\ cos\ x\\ \\ \therefore 2cos^{2}x\ -\ 1 \ =\ 2\ +\ cos\ x\\ \\ \therefore 2cos^{2}x\ -\ cos\ x - 3\ \ =\ \ 0\\ \\ \therefore \ \ (2cos\ x\ -\ 3)(cos\ x + 1)\ =\ 0\\ \\ \therefore cos\ x =\ -1 \Rightarrow \ x\ =\ 180^{\circ}\\ \\2cos\ x - 3 = 0\ has\ no\ solution$

edit: provided 180 deg allowed; i.e. x <= 180 deg

Drongoski · Apr 3, 2009

Trig and differentiate Q (1 Viewer)

sdd10

New Member

shaon0

...

Drongoski

Well-Known Member

Drongoski

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)