How do you find all real numbers in the interval [0,360) that satisfy the equation and use degree measure #4sec^2(x)tan(x)=8tan(x)#?

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Answer 1 · 2015-12-06T10:53:35

#x=45^@#, or #x=135^@#

Unless there are some typing mistake, you can simplify the tangent, as long as #x\ne 0#:

#4sec^2(x)\cancel(tan(x))=8\cancel(tan(x))#

#4sec^2(x)=8#

#sec^2(x)=2#

And since #sec(x)=1/cos(x)#, this means

#1/cos^2(x)=2 \implies cos^2(x)=1/2\implies cos(x)=\pm1/\sqrt(2)#

Which is true for #x=45^@#, and #x=135^@#