Question 1205834


{{{sin^2(x) + sin(x) = 2}}}, where {{{0 < x <2pi}}}

{{{sin^2(x) + sin(x) - 2=0}}}

{{{sin^2(x) + 2sin(x)-sin(x) - 2=0}}}

{{{(sin^2(x)-sin(x)) + (2sin(x) - 2)=0}}}

{{{sin(x)(sin(x)-1) + 2(sin(x) - 1)=0}}}

{{{(sin(x)+ 2)(sin(x) - 1)=0}}}

{{{sin(x) + 2=0}}} =>{{{sin(x) =-2}}}=> {{{x=sin^-1(-2)}}}=> no real solution

{{{sin(x) - 1=0}}} =>{{{sin(x) = 1}}}=>{{{x=sin^-1(1)}}}

=>general solution:

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

solution in given interval {{{0<x<2pi}}}:

{{{x=pi/2}}}



answer:

E. x = 𝛑/2