SOLUTION: Solve the following equations for all x, unless the domain is restricted. sin^2(x) + sin(x) = 2, where 0 < x <2𝛑 A. x = 1, -2 B. x = 2𝛑/3 C. x = 𝛑/4 D. no solutio

Algebra ->  Trigonometry-basics -> SOLUTION: Solve the following equations for all x, unless the domain is restricted. sin^2(x) + sin(x) = 2, where 0 < x <2𝛑 A. x = 1, -2 B. x = 2𝛑/3 C. x = 𝛑/4 D. no solutio      Log On


   

Question 1205834: Solve the following equations for all x, unless the domain is restricted.
sin^2(x) + sin(x) = 2, where 0 < x <2𝛑
A. x = 1, -2
B. x = 2𝛑/3
C. x = 𝛑/4
D. no solution
E. x = 𝛑/2
F. x = 𝛑/2, 4𝛑
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

sin%5E2%28x%29+%2B+sin%28x%29+=+2, where 0+%3C+x+%3C2pi
sin%5E2%28x%29+%2B+sin%28x%29+-+2=0
sin%5E2%28x%29+%2B+2sin%28x%29-sin%28x%29+-+2=0
%28sin%5E2%28x%29-sin%28x%29%29+%2B+%282sin%28x%29+-+2%29=0
sin%28x%29%28sin%28x%29-1%29+%2B+2%28sin%28x%29+-+1%29=0
%28sin%28x%29%2B+2%29%28sin%28x%29+-+1%29=0
sin%28x%29+%2B+2=0 =>sin%28x%29+=-2=> x=sin%5E-1%28-2%29=> no real solution
sin%28x%29+-+1=0 =>sin%28x%29+=+1=>x=sin%5E-1%281%29
=>general solution:
x=pi%2F2%2B2pi%2An
solution in given interval 0%3Cx%3C2pi:
x=pi%2F2


answer:
E. x = 𝛑/2

Answer by ikleyn(52937) About Me  (Show Source):
You can put this solution on YOUR website!
.


This equation is good to solve it mentally in your head, by applying reasoning. 


Indeed, sin(x) by modulus is never greater than 1.


Therefore, if sin^2(x) + sin(x) = 2, it may happen ONLY if sin(x) = 1.


But then x = pi%2F2 is the only solution in the given interval.

Solved.