allvalues of x in the interval [0,2pi]?

Question

1. 2sin^2x = 1 what formula do i use to get this to be reasonable?

2. sinx>cosx i got (pi/2,pi) from the acronym A S T C on the graph. is that correct?

Answer 1

1. You don't need one. Divide both sides by 2 to get sin^2 x = 1/2
so sin x = +- sqrt(1/2) = +- sqrt (2/4) = +- sqrt2 / 2 which happens at pi/4, 3pi/4, etc,

2. ASTC gives you the quadrants where one would be positive and the other negative, so for sure anything in Quadrant II would work. But there are also places in QI and Q III. sinx= cosx if x = pi/4 so when pi/4 < x < pi/2 it is also true. Look for what half of Q III the same is true.

Answer 2

1.)
2sin(x)^2 = 1
sin(x)^2 = (1/2)
sin(x) = sqrt(1/2)
sin(x) = (sqrt(2))/2
x = 45° or 135°

ANS : (pi/4) or ((3pi)/4)

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2.)
sin(x) > cos(x)

x < ((5pi)/4)
or
x > (pi/4)

since this is an inequality, there is a area of answers within a certain area that can be the answer.

For ex:

(3pi/4) can be an answer, since sin(3pi/4) gives you (sqrt(2)/2) and cos(3pi/4) gives you -(sqrt(2)/2)

Answer 3

1. 2sin^2x=1
sin^2x=1/2
sqr rt.= square root
sin x= sqr rt of (1/2) or 1/ sqr rt of 2
use conjugate. multiply 1/ sqr rt of 2 by
sqrt rt of 2/sqr rt of 2 <
sin x= sqr rt of 2 all over 2!!!

what radians have sine that is sqr rt of 2 all over 2??
45degrees or pi/4..and so on..

Answer 4

I don't know. BLAGH!