2007-02-04 16:21:52 · 4 answers · asked by Christie D 1 in Science & Mathematics ➔ Mathematics
Solve for x.
(tanx)(sin²x) = tanx
sin²x = 1
sinx = ±1
x = π/2 + πk, where k is an integer
2007-02-04 16:42:11 · answer #1 · answered by Northstar 7 · 0⤊ 0⤋
If the tan are equal the angles differ in k*pi
so x*sin^2x = x+k*pi .So x( 1-sin^2x) =-k*pi. But 1-sin^2x= cos^2x
Take for k 0,-1,-2......
cos^2x= -kpi The only value you can take for k is 0 because cos^2 must be >=0 ans <=1
so cos x= 0
x= pi/2+h*pi (h integer) the interval (0,2pi) you get
x=pi/2 and x= 3pi/2
2007-02-04 22:49:15 · answer #2 · answered by santmann2002 7 · 0⤊ 0⤋
No.
If this is true than the sin^2x must be 1 always.
2007-02-04 16:34:05 · answer #3 · answered by Tim L 1 · 0⤊ 0⤋
not true
tanx sin^2x
=tanx(1-cos^2x)
=(sinx/cosx)(1-cos^2x)
=(sinx/cosx - sinxcosx)
=tanx-sinxcosx
2007-02-04 16:42:36 · answer #4 · answered by grandpa 4 · 0⤊ 0⤋