if sin(x)^2-cos(x)=1/4, then the value of x between 0 and 360 are?

Answer 1

sinx^2-cosx=1/4
we know that sin^2x equals 1-cos^2x so
1-cos^2x-cosx=1/4
=>cos^2x+cosx-3/4=0

Ok,time to solve the equation:
1:cosx=1/2
2:cosx=-3/2

As we know,cosx is always variable between -1&1
So cosx=-3/4 doesn't seem to be acceptable
=>We have cosx=1/2
=>x=2kpi-pi/3
,that is,x=60',300',....
so the answers to your question,that is,x betwen 0&360 are 60&300
Hope this answer satisfies you:)

Answer 2

The correct answer is 60 and 300 degrees because both sine and cosine are postive in the first quandrant and in the fourth quadrant we know that cosine is positive and the sine is negative. However, we must remember that sine is being squared so regardless of what the sign is the result will alwayz be positive.
Therefore
sin(60)^2 - cos(60)=1/4
[sqrt(3)/2]^2-1/2=1/4
3/4-1/2=1/4
1/4=1/4

and

sin(300)^2 - cos(300)=1/4
[-sqrt(3)/2]^2-1/2=1/4
3/4-1/2=1/4
1/4=1/4

Answer 3

Sin(x)^2 or sin^2(x)
If it is sin^2(x) then sin^2(x)= 1-cos^2(x)
the equation then be came
1- t^2 -t = 1/4 with t = cos(x) , -1 <=t <=1,
we then have t= 1/2( in range of cosin value) and t= -3/2 (not a cosin value)
cos(x)= 1/2 then x= 60 or x= 300

Answer 4

How did X get between 0 and 360?

Answer 5

X= 60.

(sin 60)^2 = 0.75
cos 60 = 0.5

0.75 - 0.5 = 0.25 = 1/4

Answer 6

Answer: I'm a ninja?

Answer 7

no one really knows.....now take off your pants