Thread: Please help.Trignometric equations.

Can someone help me on these

1.

2.

3.

4.


thanks

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

Originally Posted by hacker804

Can someone help me on these

1.

2.

3.

4.


thanks

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.

√3 tan⁡x-sec⁡x-1=0
√3 sin⁡x/cos⁡x -1/cos⁡x -1=0
√3 sin⁡x-1- cos⁡x=0
√3/2 sin⁡x-1/2 cos⁡x=1/2 {2 is √((√3)^2+(-1)^2 )}
cos⁡〖π/6 sin⁡x-〗 sin⁡〖π/6〗 cos⁡x=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

sec⁡θ=sec⁡3θ
cos⁡θ=cos⁡3θ
cos⁡θ=4cos^3⁡θ-3 cos⁡θ
0=4 cos^3⁡θ-4 cos⁡θ
4 cos⁡θ {cos^2⁡θ-1}=0
cos⁡θ (cos⁡θ-1)(cos⁡θ+1)=0
cos⁡θ=0 or cos⁡θ=1 or cos⁡θ=-1
cos⁡θ=cos⁡0 or cos⁡θ=cos⁡〖π/2〗 or cos⁡θ=cos⁡((-π)/2)