solve for x:sinxtan2x=sinx?

4 ⤊

solve for x:sinxtan2x=sinx?

where 0<\x<\2pi

2007-11-24 17:22:07 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics

4 answers

sin x(tan 2x - 1) = 0
sin x = 0 , tan 2x = 1

Solutions for sin x = 0
x = 0 , π , 2π

Solutions for tan 2x = 1
2x = π / 4 , 5π / 4 , 9π / 4 , 13π / 4
x = π / 8 , 5π / 8 , 9π / 8 , 13π / 8

2007-11-25 06:14:15 · answer #1 · answered by Como 7 · 3⤊ 1⤋

sinxtan2x = sinx
=> sinx(tan2x -- 1) = 0
when sinx = 0, no value of x between 0 and 2pi.
when tan2x -- 1 = 0
=> tan2x = 1 = tan pi/4
=> 2x = npi + pi/4
=> x = npi/2 + pi/8
x = pi/8, 5pi/8.

2007-11-24 18:09:41 · answer #2 · answered by sv 7 · 0⤊ 0⤋

x = 0, 5pi/8, pi, 13 pi/8

2007-11-24 17:33:07 · answer #3 · answered by sayamiam 6 · 1⤊ 0⤋

x= 22.5 degrees or pi/8

that makes tan2x=1

2007-11-24 17:29:46 · answer #4 · answered by Anonymous · 0⤊ 1⤋