Question 1192743


{{{cos(2x)-2sin(x)-cos^2 (x) = -3}}} on the interval {{{x}}} ∈ [{{{0}}},{{{2π}}}]


since {{{cos(2x)=cos^2(x) - sin^2(x)}}}, we have


{{{cos^2(x) - sin^2(x)-2sin(x)-cos^2 (x) = -3 }}}

 {{{-sin^2(x)-2sin(x)= -3 }}}

 {{{-3sin^2(x)= -3}}} 

{{{ sin^2(x)= -3/-3 }}}

{{{ sin^2(x)= 1}}}

{{{ sin(x)= sqrt(1)}}}

{{{sin(x)= 1}}}


{{{x=sin^-1(1)}}}

{{{x=pi/2}}}


periodicity of sin is {{{2pi}}}

general solution:

{{{x=pi/2+2pi*n}}}


on the interval {{{x}}} ∈ [{{{0}}},{{{2π}}}] : {{{x=pi/2}}}