Tan2x+secx=1/2

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Tan2x+secx=1/2

Helb

>>

Hello,

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

Cordially,

Anon

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>>7781173

What did I do wrong.

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

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

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>>7781168

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

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>>7781168

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

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>>7781252

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

Also where would I go from

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

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>>7781168

Use(sin2x/cos2x) = tan 2x. and (1/cosx) = secx .

Then use the double angel identities for sin2x and cos2x.

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>>7781295

Sorry guys the problem actually is

tan2x + sec2x = 0.5

From this I got to..

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

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sin2x +1 = 0.5 cos2x

you can solve this right?

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>>7781309

factor.

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just use euler

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>>7781414

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

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

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

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

>>7781416

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>>7781168

OP here

Tan2x+1/cos2x=1/2

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

Stuck on this point,Anyone?

Thread images: 1

Thread DB ID: 418623

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