Mathematics, 29.05.2020 03:58 bigblow

To help solve the trigonometric inequality 2sin'(x)+ cos(2x)-2<-1, Tracey successfully graphed the equations
y=2sinº(*)+ COS(2x)-2 and Y--1 with her graphing calculator. Which of the following statements is true about the
inequality?
There is no solution because even though you cannot see it, the graph of the equation y-2sin? (x)+cos(2x)-2 is above
the graph of the equation Y--1 for all values of x.
There is no solution because the graph of the equation y-2sin(x)+cos(2x)- 2 is the same as the graph of the equation
y=-1, so y=2sin?(x)+cos(2x)– 2 is never belowy--1.
There are an infinite number of solutions because even though you cannot see it, the graph of the equation
y=2sin'(x)+ COS(2x)- 2 is below the graph of the equation Y--1 for all values of x.
There are an infinite number of solutions because the graph of the equation y-2sin(x)+ cos(2x)-2 is the same as the
graph of the equation Y--1, so y-Zsinº($)+COS(2x)-2 is never below Y--1

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