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How do you solve #cos2x=secx#?

1 Answer

Apr 24, 2015

#cos(2x) = 2cos^2(x) - 1#

#2cos^2(x)-1 = 1/cos(x)#

Multiply by cos(x) both side :

#2cos^3(x)-cos(x) = 1#

#=>2cos^3(x) - cos(x) - 1 = 0#

Let's #cos(x) = X#

#=> 2X^3-X-1=0#

We see one solution #X=1#

#cos(x) = 1 => cos(x) = cos(0) => x = 0 + 2kpi# with #k in ZZ#

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