trig probs: sin(x)- 2sin^2(x)= 0, sinx= cos2x, tanx + secx =

[velocity]

Can you make sure I did these right? If not can you tell me where I went wrong? Thanks.

1) sin(x)- 2sin^2(x)= 0
I factored out sinx to get sinx (1-2sinx)=0
So that gives me sinx =0
and (1-2sinx)=0 --- sinx= 1/2

2) sinx= cos2x
Divide by cosx to get sinx/ cosx = 2
Then i look on unit circle and see that pi/4 and 5pi/4 match this

3) tanx + secx =sqare root (3)
I changed it to read (sinx/cosx) + (1/cosx) = the square root of (3)
This simplifies to sinx/ cosx = the square root of (3)

 

[skeeter]

Re: trig prob

Can you make sure I did these right? If not can you tell me where I went wrong? Thanks.

1)sin(x)- 2sin^2(x)= 0
I factored out sinx to get sinx (1-2sinx)=0
So that gives me sinx =0
and (1-2sinx)=0 --- sinx= 1/2
ok ... so x = ?

2) sinx= cos2x
Divide by cosx to get sinx/ cosx = 2
no ... sinx = 1 - 2sin²x
2sin²x + sinx - 1 = 0
(2sinx - 1)(sinx + 1) = 0
proceed as before ...


3)tanx + secx =sqare root (3)
I changed it to read (sinx/cosx) + (1/cosx) = the square root of (3)
This simplifies to sinx/ cosx = the square root of (3)
no ... it simplifies to
(sinx + 1)/cosx = sqrt(3)
square both sides ...
(sinx + 1)²/cos²x = 3
(sinx + 1)²/(1 - sin²x) = 3
(sinx + 1)²/[(1 - sinx)(1 + sinx)] = 3
(sinx + 1)/(1 - sinx) = 3
sinx + 1 = 3 - 3sinx
4sinx = 2
sinx = 1/2
x = ?