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I must be having a birdbrain day today.

Question: Find the values of x in the interval 0<x<360 for which Sin2x + Cos2x =0

I have attempted to square and then sub in using the identity: sin^2x + cos^2x = 1, but this just ends with 1 as both sin and cos just cancel each other out, and I don’t see a clear way of using tan x=sin x/ cos x .

Is there something that I am missing?

Question: Find the values of x in the interval 0<x<360 for which Sin2x + Cos2x =0

I have attempted to square and then sub in using the identity: sin^2x + cos^2x = 1, but this just ends with 1 as both sin and cos just cancel each other out, and I don’t see a clear way of using tan x=sin x/ cos x .

Is there something that I am missing?

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(Original post by

Try and rearrange to the form “tan 2x= ?”

**JGLM**)Try and rearrange to the form “tan 2x= ?”

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#5

Find the values of x in the interval 0<x<360 for which Sin2x + Cos2x =0

cos2x = -sin2x

-tan2x =1, 1+tan2x =0

0<x<360

0<2x<720

tan0<tan2x<tan720

then it may get the answer

cos2x = -sin2x

-tan2x =1, 1+tan2x =0

0<x<360

0<2x<720

tan0<tan2x<tan720

then it may get the answer

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(Original post by

Use some identities for cos2x and sin2x?

**laurawatt**)Use some identities for cos2x and sin2x?

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(Original post by

Find the values of x in the interval 0<x<360 for which Sin2x + Cos2x =0

cos2x = -sin2x

-tan2x =1, 1+tan2x =0

0<x<360

0<2x<720

tan0<tan2x<tan720

then it may get the answer

**deskochan**)Find the values of x in the interval 0<x<360 for which Sin2x + Cos2x =0

cos2x = -sin2x

-tan2x =1, 1+tan2x =0

0<x<360

0<2x<720

tan0<tan2x<tan720

then it may get the answer

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#8

(Original post by

Can you show me how you took cos2x=-sin2x to be -tan2x=1? ? I am only aware of the identity tan x=sin x/cos x. I must be missing something

**KingRich**)Can you show me how you took cos2x=-sin2x to be -tan2x=1? ? I am only aware of the identity tan x=sin x/cos x. I must be missing something

Last edited by deskochan; 1 month ago

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#9

Identities mentioned earlier

Sin2x = 2sinxcosx

Cos2x = cosx^2 - sinx^2

Cos2x = 2cosx^2 - 1

Cos2x= 1-2sinx^2

Sin2x = 2sinxcosx

Cos2x = cosx^2 - sinx^2

Cos2x = 2cosx^2 - 1

Cos2x= 1-2sinx^2

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#10

(Original post by

Find the values of x in the interval 0<x<360 for which Sin2x + Cos2x =0

cos2x = -sin2x

**deskochan**)Find the values of x in the interval 0<x<360 for which Sin2x + Cos2x =0

cos2x = -sin2x

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(Original post by

cos2x = -sin2x (divided both sides by cos2x)

**deskochan**)cos2x = -sin2x (divided both sides by cos2x)

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#13

(Original post by

Okay, I see what you did. According to the book I am working from the first x value is 67.5 degrees, therefore the solution above doesn't work out as it gives an x value of -22.5

**KingRich**)Okay, I see what you did. According to the book I am working from the first x value is 67.5 degrees, therefore the solution above doesn't work out as it gives an x value of -22.5

Last edited by deskochan; 1 month ago

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(Original post by

In the case of the negative value of the tangent region, we can consider two quadrants (II and IV) such that quadrant II (sine region) is 90+(-22.5) = 67.5 and quadrant IV (cosine region) is 360+(-22.5) = 337.5.

**deskochan**)In the case of the negative value of the tangent region, we can consider two quadrants (II and IV) such that quadrant II (sine region) is 90+(-22.5) = 67.5 and quadrant IV (cosine region) is 360+(-22.5) = 337.5.

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#15

(Original post by

Superstar! I need to stop for the day because this seems so simple now that you have explained it. Thank you for this

**KingRich**)Superstar! I need to stop for the day because this seems so simple now that you have explained it. Thank you for this

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