How to solve cos2x<3sinx + 2 for (-π<x<π)?

2 Answers
Jul 23, 2017

cos 2x < 3sin x +2
Replace cos 2x by #(1 - 2sin^2 x)# -->
#1 - 2sin^2 x < 3sin x + 2#
#f(x) = 2sin^2 x + 3sin x + 1 > 0#.
First, solve f(x) = 0.
Since a - b + c = 0, use shortcut. The 2 real roots are:
sin x = - 1 and #sin x = -c/a = - 1/2#
When sin x = - 1 --> #x = - pi/2#
When #sin x = -1/2# --> #x = - pi/6# and #x = -(5pi)/6#
Within the interval# (-pi, 0)#, --> f(x) = 0 when :
#x = -(5pi)/6, x = (-pi)/2, and x = (-pi/6)#;
Within interval (0, pi), --> f(x) always positive because sin x > 0.
Answer: f(x) > 0 within the 3 intervals:
#(- pi, - (5pi)/6) and (-pi/6, 0), and (0, pi)#
We can check the answer by using graphing calculator to graph the function f(x). The parts of the graph that stay above the x-axis give the answers for f(x) > 0.
Check by calculator:
#f(-160^@) = 2(-0.34)^2 + 3(-0.34) + 1 = 0.23 - 1.02 + 1 > 0#. OK
#f(- 10^@) = 2(-0.17)^2 - 3(0.17) + 1 = 0.06 - 0.52 + 1 > 0#. OK

Jul 24, 2017

Given: #cos(2x)<3sin(x) + 2; -pi < x < pi#

Substitute #1 - 2sin^2(x)# for #cos(2x)#:

#1 - 2sin^2(x)<3sin(x) + 2; -pi < x < pi#

Add #2sin^2(x) - 1# to both sides:

#0 < 2sin^2(x)+3sin(x) + 1; -pi < x < pi#

Please look at #y = 2sin^2(x)+3sin(x) + 1; -pi < x < pi# and observe where 0 is less than the y value:

www.Desmos.com/calculator

I have marked the points #(-5pi/6,0)# and #(-pi/6,0)#. Between these two points, the graph is less than or equal to 0; everywhere else 0 is less than the graph as the inequality specifies.

We can show this analytically by factoring the quadratic:

#(2sin(x)+1)(sin(x)+1)#

This gives us the roots:

#sin(x) = -1/2# and #sin(x) = -1#

The two points that I have marked, #(-5pi/6,0)# and #(-pi/6,0)#, correspond to the first root.

#x = sin^-1(-1/2)#

#x = -5pi/6 and -pi/6#

The second root is a false solution, because it does not cross the #y = 0# line.

Therefore, the solution to the inequality is these two regions:

#-pi < x < -5pi/6# and #-pi/6 < x < pi#