Answer:
x = {π/4, 7π/6, 5π/4, 11π/6} +2kπ . . . for any integer k
Step-by-step explanation:
[tex]\dfrac{\sin^3{x}}{1+\cos{x}}+\dfrac{\cos^3{x}}{1+\sin{x}}=\cos{2x}+2\cos{x}-1\\\\\dfrac{\sin{x}(1-\cos^2{x})}{1+\cos{x}}+\dfrac{\cos{x}(1-\sin^2{x})}{1+\sin{x}}=\cos^2{x}-\sin^2{x}+2\cos{x}-1\\\\\sin{x}(1-\cos{x})+\cos{x}(1-\sin{x})=\cos^2{x}-\sin^2{x}+2\cos{x}-1\\\\\sin{x}+\cos{x}-2\sin{x}\cos{x}=2\cos{x}-2\sin^2{x}\qquad\text{use $1=s^2+c^2$}\\\\\sin{x}+2\sin^2{x}-\cos{x}-2\sin{x}\cos{x}=0\\\\\sin{x}(1+2\sin{x})-\cos{x}(1+2\sin{x})=0\\\\(\sin{x}-\cos{x})(1+2\sin{x})=0[/tex]
This will have solutions where the factors are zero.
sin(x) -cos(x) = 0
Dividing by cos(x), we have ...
tan(x) -1 = 0
x = arctan(1) = π/4, 5π/4
1 +2sin(x) = 0
sin(x) = -1/2
x = arcsin(-1/2) = 7π/6, 11π/6
The four solutions in the interval [0, 2π] are x = {π/4, 7π/6, 5π/4, 11π/6}. Solutions repeat every 2π radians.
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Additional comment
We have made use of the factoring of the difference of squares:
(1 -a^2) = (1 -a)(1 +a)
and we have made use of the cosine double angle identity:
cos(2x) = cos(x)^2 -sin(x)^2
The "Pythagorean" identity for sine and cosine was used several times:
1 = sin(x)^2 +cos(x)^2
