Hello,

It appears that you are in Trig Identity Hell. Stay there for a bit. There is much to learn.

Cordially,
Anon

>>7781173
What did I do wrong.

I've had a another try and got to ...

4sinxcosx +1 +2sin^2x=0

>>7781168
Remember Sin(2X) = 2sin(X)cos(X)

>>7781168
And sin^2(X) = (1+cos(2x))/2

>>7781252
I've got sin^2x=(1-cosx)/2 [not as +cosx]

Also where would I go from
4sinxcosx +1 +2sin^2x=0

>>7781168
Use(sin2x/cos2x) = tan 2x. and (1/cosx) = secx .
Then use the double angel identities for sin2x and cos2x.

>>7781295
Sorry guys the problem actually is

tan2x + sec2x = 0.5

From this I got to..

4sinxcosx + 1 +2sin^2x=0 what from here?

sin2x +1 = 0.5 cos2x

you can solve this right?

>>7781309
factor.

just use euler

>>7781414

2sinx (sinx+2cosx) +1 =0

Is one of the solutions -30 degrees ? If not how do I get the solution from this

>2sinx (sinx+2cosx) +1 =0
>Is one of the solutions -30 degrees ? If not how do I get the solution from this
>>7781416

>>7781168
OP here

Tan2x+1/cos2x=1/2


2sinx (sinx+2cosx)+1=0

Stuck on this point,Anyone?