What is the solution for the equation sinx=cosx?
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sin x - cos x=0
We could divide the equation by (cos x) and the result will be:
sin x /cos x - 1 =0
But we know that the ratio between sinx and cos x determine the tangent function.
tg x -1 =0
We'll move the free term in the right side of the equal sign, like this:
tg x = 1
x= arctg 1 + k*pi
x = pi/4 + k*pi
We could also make the note that if the ratio sin x / cos x=1, it means that the terms from the ratio are equal
=> sin x = cosx!
If sin x = cos x => the angles of the triangle are equal, too, so, in a right angle triangle, the angles could only be = 45 degrees (the conclusion is based on fact that in a triangle, the sum of angles is 180 degrees, and one of them is 90 degrees and the other 2 are equals, so
90 + 2*x=180
sinx = cosx , given. square both sides and get
(sinx)^2 = ( cosx)^2 =1-(sinx)^2x , as (sinx)^2+cosx)^2 = 1 is a well known identiy .
Rearrange to get, 2(sinx)^2 = 1
Therefore, sinx = +1/sqrt2 or -1/sqrt2
When sinx =1/sqrt2 ,x=45degree or
when sinx -1/sqrt2, x=180+45 = 225 degree.
cos45 = 1/sqrt2 =sin45 and
co225 = -1/sqrt2 =sign225
so X=pi/4(4n+1)..................n belongs to Real numbers
n=0 you get X=45degrees
n=1 you get X=225degrees
put n=-1,-2,-3,....2,3,4,............you got perfact answers
Square both sides of the original equation to get
From the identity `cos^2x=1-(sin^2x)`
we get `sin^2x=1-(sin^2x)`
Rearrange to get `2(sin^2x)=1`
`sinx=1/sqrt(2) or sinx=-1/sqrt(2)`
When `sinx=1/sqrt(2)` ,`x=45^@ or pi/4 rad`
When `sin=-1/sqrt(2)` , `x=225^@ or 5pi/4 rad`
sinx = cosx
tanx = 1
Taking inverse of tan on both sides,
x = 45°
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