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3tanx = 8/sinx

can be written as

3cos^2x + 8cosx -3 = 0

HOW??

 Apr 12, 2015

Best Answer 

 #1
avatar+27529 
+10

Multiply both sides by sinx

3tanx*sinx = 8

 

tanx = sinx/cosx so the above can be written as

3*sin2x/cosx = 8

 

Multiply both sides by cos x

3*sin2x = 8cosx

 

sin2x = 1 - cos2x so the above becomes

3*(1 - cos2x) = 8cosx

3 - 3cos2x = 8cosx

 

Add 3cos2x to both sides of the equation

3 = 3cos2x + 8cosx

 

Subtract 3 from both sides of the equation

0 = 3cos2x + 8cosx - 3

.

 Apr 12, 2015
 #1
avatar+27529 
+10
Best Answer

Multiply both sides by sinx

3tanx*sinx = 8

 

tanx = sinx/cosx so the above can be written as

3*sin2x/cosx = 8

 

Multiply both sides by cos x

3*sin2x = 8cosx

 

sin2x = 1 - cos2x so the above becomes

3*(1 - cos2x) = 8cosx

3 - 3cos2x = 8cosx

 

Add 3cos2x to both sides of the equation

3 = 3cos2x + 8cosx

 

Subtract 3 from both sides of the equation

0 = 3cos2x + 8cosx - 3

.

Alan Apr 12, 2015

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