krypticlemonjuice
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- 2023
Someone please explain how to do this question
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Trial Q (1 Viewer)
- Thread starter krypticlemonjuice
- Start date
Someone please explain how to do this question
synthesisFR
afterhscivemostlybeentrollingdonttakeitsrsly
Is it A?
yeah, how’d you get it?
synthesisFR
afterhscivemostlybeentrollingdonttakeitsrsly
okay move the two to the left but it doesn't matter
so u have to see if any solutions are equal to zero
if u look at the equation, its a periodic function since both terms have sin or cos
so then if it ever crosses the x axis it will do so forever
but they also limited the domain from 0 to 2pi
if u quickly check those two endpoints, u will see that
they are both at a max the final answer u get on the LHS is 1, which occurs at 0 and 2pi
but u don't know if it goes below the x axis
if u test x=pi/4 which is when they r both 1/sqrt2, u get like 2sqrt2-2 which is still greater than one
after that the values for each x go opposite to each other idk how to explain this bit but yeah it made sense it my head
or u can just quickly test the next value and u see that the functions value increases
so its always above the x axis and hence there is no solution
so u have to see if any solutions are equal to zero
if u look at the equation, its a periodic function since both terms have sin or cos
so then if it ever crosses the x axis it will do so forever
but they also limited the domain from 0 to 2pi
if u quickly check those two endpoints, u will see that
they are both at a max the final answer u get on the LHS is 1, which occurs at 0 and 2pi
but u don't know if it goes below the x axis
if u test x=pi/4 which is when they r both 1/sqrt2, u get like 2sqrt2-2 which is still greater than one
after that the values for each x go opposite to each other idk how to explain this bit but yeah it made sense it my head
or u can just quickly test the next value and u see that the functions value increases
so its always above the x axis and hence there is no solution
No solutions because:
The smallest each of the terms:
can be is 1. Therefore the sum of these 2 terms cannot be less than 2. But this sum equals 2 only when each term equals 1 simultaneously. Each term equals 1 only when 
; but this cannot happen simultaneously, e.g. when 
The smallest each of the terms:
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