Can someone help me on these

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thanks

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- August 15th 2012, 05:05 AMhacker804Please help.Trignometric equations.
Can someone help me on these

1.

2.

3.

4.

thanks - August 15th 2012, 04:33 PMAmerRe: Please help.Trignometric equations.
first one 2x+3x = 90, or pi/2. and same angle in the third quarter with the 2n pi

the seond one use the identity

tan^2 x + 1 = sec ^2 x - August 15th 2012, 05:00 PMProve ItRe: Please help.Trignometric equations.
1.

Write to get the polynomial .

It's easy to see that satisfies this equation, so is a factor, and long division gives

So you need to solve (easy), as well as and (not so easy). All will have exact value answers though. This will help you. - August 17th 2012, 08:45 AMsujith123Re: Please help.Trignometric equations.
√3 tanx-secx-1=0

√3 sinx/cosx -1/cosx -1=0

√3 sinx-1- cosx=0

√3/2 sinx-1/2 cosx=1/2 {2 is √((√3)^2+(-1)^2 )}

cos〖π/6 sinx-〗 sin〖π/6〗 cosx=sin〖π/6〗

sin〖(π/6-x)=〗 sin〖π/6〗

π/6-x=nπ+(-1)^n π/6

x=π/6-nπ-(-1)^n π/6

here n is a integer - August 17th 2012, 08:55 AMsujith123Re: Please help.Trignometric equations.
secθ=sec3θ

cosθ=cos3θ

cosθ=4cos^3θ-3 cosθ

0=4 cos^3θ-4 cosθ

4 cosθ {cos^2θ-1}=0

cosθ (cosθ-1)(cosθ+1)=0

cosθ=0 or cosθ=1 or cosθ=-1

cosθ=cos0 or cosθ=cos〖π/2〗 or cosθ=cos((-π)/2)