trig question (1 Viewer)

[pLuvia]

find the general solution to the following equations

1. cos (theta)/2 = 1
2. 2 sinx = secx
3. sec2x = cosec2x

 

[Kutay]

i can do the second one not sure if right though

2sinx = secx

2sinx/cosx = 1/cosx

gets 2sincosx = 1

gets Sin2@ = 1

not sure from here

sin x = k and a = sin^-1 k

then x = nPI + (-1)^n a

i know that what will look like but not sure because it becomes a 2@ need to work on these but i know that what it turn ot to sometinhg like that

 

[gman03]

i can do the second one not sure if right though

2sinx = secx

2sinx/cosx = 1/cosx

Sec x = 1 / cosx, so

2sinx = 1 / cos x, so

1 = 2sinx cosx = sin2x, so

2x = Pi/2 + 2kPi, for integer k

Therefore x = Pi/4 * ( 1 + 4k), for integer k

 

[gman03]

3. sec2x = cosec2x

sec2x = cosec2x, that is

1/cos2x = 1/sin2x, cos2x =/= 0 and sin2x =/= 0

On top of that, tan 2x = 1 (by multiplying each side by sin2x)

So the solutions are among 2x = Pi/4 + kPi (for integer k)

Check: cos2x = cos (Pi/4 + kPi) and sin2x = cos (Pi/4 + kPi) can't be 0.

Therefore the solutions are x = Pi/8 * (1 + 4k), for integer k

 

[pLuvia]

Sec x = 1 / cosx, so

2sinx = 1 / cos x, so

1 = 2sinx cosx = sin2x, so

2x = Pi/2 + 2kPi, for integer k

Therefore x = Pi/4 * ( 1 + 4k), for integer k

the book answer is nPI/2 * (-1)^n * PI/4

 

[BoganBoy]

You probably found the answer already since its a long time after u posted. About the second question which the guys upstairs were trying to do:

2sinx=secx
=1/cosx
2sinxcosx =1
sin2x=1

therefore, six2x=sin(pi/2)

therefore, using the equation: x=nPi+(-1)^n(a), the general solution is:

2x=nPi+(-1)^n(Pi/2), therefore, x=nPi/2+(-1)^nPi/4

 

[BoganBoy]

3. sec2x = cosec2x


1/cos2x=1/sin2x

sin2x/cos2x=1

tan2x=1=tan(pi/4)

Therefore, 2x=nPi+Pi/4

x=nPi/2+Pi/8