Question 1205304


 {{{cos(2x) + sin^2(x) = 4/9}}} .... where {{{0 <= x <= 2pi}}}


{{{cos(2x) + sin^2(x) = 4/9}}}

use identity{{{ cos(2x)=cos^2(x) - sin^2(x)}}}

{{{cos^2(x) - sin^2(x) + sin^2(x) = 4/9}}}

{{{cos^2(x) = 4/9}}}

{{{cos(x) =sqrt (4/9)}}}

{{{cos(x) =2/3}}}


solutions in given interval:

{{{x=cos^-1(2/3)}}}

{{{x=0.84106867}}}

{{{x=48.19}}}°


{{{x=cos^-1(-2/3)}}}

{{{x=2.3005239}}}

{{{x=131.8}}}°


{{{x=2pi-cos^-1(2/3)}}}

{{{x=5.4421166}}}

{{{x=311.81}}}°


{{{x=2pi-cos^-1(-2/3)}}}

{{{x=3.9826613241}}}

{{{x=228.19}}}°