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solve the inequalities tan X> cos X

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appreciate0831 | Student | (Level 1) Honors

Posted July 9, 2012 at 1:57 PM via web

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solve the inequalities

tan X> cos X

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted July 9, 2012 at 6:22 PM (Answer #2)

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Solve `tan(x)>cos(x)` :

Then` ` `(sinx)/(cosx)>cosx` . There are two cases:

` `(1) `cosx>0` :

`sinx>cos^2x`

`sinx>1-sin^2x`

`sin^2x+sinx-1>0`

`sinx=(-1+-sqrt(1-4(1)(-1)))/2`

`sinx=1/2(-1+-sqrt(5))`

`x=sin^(-1)(1/2(-1+sqrt(5))~~.666` . The other zero does not work.

`cosx>0 ==> 2npi +-pi/2<x<2npi +pi/2`

So `2npi +.666<x<2npi + pi/2`

(2) `cosx<0`

`sinx<cos^2x`  (Multiplied by a negative)

`sinx<1-sin^2x`

The same zero occurs. (This is where `tanx=cosx` . These points are symmetric about the lines `x=pi/2 +2npi` . Since 1 crossing is at `x~~.66623 + 2npi` , the other crossing is at `x~~2.47535+2npi` )

`cosx<0==> 2npi + pi/2<x<2npi+(3pi)/2`

Thus `2npi +2.475<x<2npi + (3pi)/2`

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The exact solutions are:

`2pin + sin^(-1)((-1+sqrt(5))/2)<x<2pin +pi/2`

`2pin+(pi-sin^(-1)((-1+sqrt(5))/2))<x<2npi+(3pi)/2`

Approximate solutions:

`2pin+.666<x<2pin+pi/2`

`2pin+2.475<x<2pin+(3pi)/2`

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