Question #3c29b

3 Answers
Dec 9, 2017

is it 1-tan^2x instead of 1-tanx?

Explanation:

1/cos^2x = 1/(1-sin^2x)
1/(1-sin^2x) = 1-1/sin^2x
1-1/(1-cos^2x) = 1-1-sec^2x
1-1-sec^2x = sec^2x = 1-tan^2x

Dec 9, 2017

sinx=0 or sinx+cosx=0

x=0, (3pi)/4, pi, (7pi)/4

Explanation:

1/cos^2x=1-tanx

1=cos^2x(1-tanx)

1=cos^2x-tanxcos^2x

1=cos^2x-sinx/cosxcos^2x

1=cos^2x-sinxcosx

sin^2x+cos^2x=cos^2x-sinxcosx

sin^2x=-sinxcosx

sin^2x+sinxcosx=0

sinx(sinx+cosx)=0

sinx=0 or sinx+cosx=0
sinx=0 or sinx=-cosx

Edit: from the comments:

For the range 0<=x<=2pi

a) sinx=0
:.x=0, x=pi

b) sinx+cosx=0
We are going to divide both sides by cosx. It is necessary to check beforehand that we are not dividing by zero. However, if cosx=0, sinx=1, and the sum is not zero. In this case, we are safe to divide by cosx/

sinx/cosx+cosx/cosx=0
tanx+1=0
tanx=-1

x=(3pi)/4, (7pi)/4

Putting this together:

x=0, (3pi)/4, pi, (7pi)/4

Dec 9, 2017

tanx=0 and tanx=-1

Explanation:

1/cos^2x=1-tanx

Identities:

color(red)(1/cos^2x=sec^2x)

color(red)(sec^2x=1+tan^2x)

So we have:

1+tan^2x=1-tanx

Subtract 1 from both sides:

tan^2x=-tanx

tan^2x+tanx=0

Factor:

tanx(tanx+1)=0

tanx=0 and tanx=-1

You haven't given an interval, so for 0<=x<=2pi

0, pi , (3pi)/4,(7pi)/4

Graph:

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