p2 trig question

solve 2sin2x = tanx

between 0 \< x \< 360

ThugzMansion7

solve 2sin2x = tanx

between 0 \< x \< 360

2sin2x = 2[2sinxcosx] = 4sinxcosx = tanx
4sinxcosx = sinx/cosx
4sinx(cosx)^2 = sinx
4sinx(cosx)^2 - sinx = 0
sinx[4(cosx)^2 - 1] = 0
sinx=0 or 4(cosx)^2 = 1
sinx=0 or cosx = 1/2
x=0, 180, 360, 60, 300.

Reply 2

RichE

sinx=0 or 4(cosx)^2 = 1
sinx=0 or cosx = 1/2 or cosx = -1/2

Reply 3

ThugzMansion7

Gaz031

cheers

Reply 4

ThugzMansion7

what about:

using cos(A + B) to prove cos2A ≡ 2cos^2 + 1

if someone could point me in the right direction...

Reply 5

C4>O7

use sin^2(A) + cos^2(A) = 1

Reply 6

username9816

ThugzMansion7

what about:

using cos(A + B) to prove cos2A ≡ 2cos^2 + 1

if someone could point me in the right direction...

cos2A = cos(A + A) = cosAcosA - sinAsinA = cos^2A - sin^2A = cos^2A - (1 - cos^2A) = 2cos^2A - 1

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