Solve the equation sin2x+cos2x=1, if 0
Solution
We\'ll recall the double angle identities for sin 2x and cos 2x:
sin 2x = 2 sinx*cos x
cos 2x = [cos^(2)] x - [sin^(2)] x
We\'ll recall the Pythagorean identity:
[sin^(2)] x + [cos^(2)] x = 1
We\'ll re-write the equation in terms of sin x and cos x:
2sin x*cos x + [cos^(2)] x - [sin^(2)] x = [sin^(2)] x + [cos^(2)] x
We\'ll remove like terms:
2 sin x*cos x - 2 [sin^(2)] x = 0
We\'ll factorize by sin x:
sin x*(cos x - sin x) = 0
We\'ll cancel each factor:
sin x = 0
x = [pi] (we\'ll exclude the values 0 and 2 [pi] )
We\'ll cancel the next factor:
cos x - sin x = 0
-tan x = -1
tan x = 1
The tangent function has positive values within the 1st and the 3rd quadrants, therefore the
values of x are:
x = [pi] /4
x = [pi] + [pi] /4
x = 5 [pi] /4
Therefore, the solutions of the equation, over the interval (0,2 [pi] ), are: { [pi] /4 ; [pi] ; 5 [pi]
/4}..
Solve the equation sin2x+cos2x=1, if 0x2piSolutionWell re.pdf
1. Solve the equation sin2x+cos2x=1, if 0
Solution
We'll recall the double angle identities for sin 2x and cos 2x:
sin 2x = 2 sinx*cos x
cos 2x = [cos^(2)] x - [sin^(2)] x
We'll recall the Pythagorean identity:
[sin^(2)] x + [cos^(2)] x = 1
We'll re-write the equation in terms of sin x and cos x:
2sin x*cos x + [cos^(2)] x - [sin^(2)] x = [sin^(2)] x + [cos^(2)] x
We'll remove like terms:
2 sin x*cos x - 2 [sin^(2)] x = 0
We'll factorize by sin x:
sin x*(cos x - sin x) = 0
We'll cancel each factor:
sin x = 0
x = [pi] (we'll exclude the values 0 and 2 [pi] )
We'll cancel the next factor:
cos x - sin x = 0
-tan x = -1
tan x = 1
The tangent function has positive values within the 1st and the 3rd quadrants, therefore the
values of x are:
x = [pi] /4
x = [pi] + [pi] /4
x = 5 [pi] /4
Therefore, the solutions of the equation, over the interval (0,2 [pi] ), are: { [pi] /4 ; [pi] ; 5 [pi]
/4}.
![Solve the equation sin2x+cos2x=1, if 0
Solution
We'll recall the double angle identities for sin 2x and cos 2x:
sin 2x = 2 sinx*cos x
cos 2x = [cos^(2)] x - [sin^(2)] x
We'll recall the Pythagorean identity:
[sin^(2)] x + [cos^(2)] x = 1
We'll re-write the equation in terms of sin x and cos x:
2sin x*cos x + [cos^(2)] x - [sin^(2)] x = [sin^(2)] x + [cos^(2)] x
We'll remove like terms:
2 sin x*cos x - 2 [sin^(2)] x = 0
We'll factorize by sin x:
sin x*(cos x - sin x) = 0
We'll cancel each factor:
sin x = 0
x = [pi] (we'll exclude the values 0 and 2 [pi] )
We'll cancel the next factor:
cos x - sin x = 0
-tan x = -1
tan x = 1
The tangent function has positive values within the 1st and the 3rd quadrants, therefore the
values of x are:
x = [pi] /4
x = [pi] + [pi] /4
x = 5 [pi] /4
Therefore, the solutions of the equation, over the interval (0,2 [pi] ), are: { [pi] /4 ; [pi] ; 5 [pi]
/4}.](https://image.slidesharecdn.com/solvetheequationsin2xcos2x1if0x2pisolutionwellre-230327113714-da69ca60/85/Solve-the-equation-sin2x-cos2x-1-if-0x2piSolutionWell-re-pdf-1-320.jpg)
