Question

Solve the equation 3), x=???? Please help me!!!

Answer 1

x = {π/4, 7π/6, 5π/4, 11π/6} +2kπ . . . for any integer k

Explanation:

`\frac{\sin^3{x}}{1+cos(x)}+\frac{\cos^3{x}}{1+sin(x)}=cos(2x)+2cos(x)-1\\\\\frac{sin(x)(1-\cos^2{x})}{1+cos(x)}+\frac{cos(x)(1-\sin^2{x})}{1+sin(x)}=\cos^2{x}-\sin^2{x}+2cos(x)-1\\\\sin(x)(1-cos(x))+cos(x)(1-sin(x))=\cos^2{x}-\sin^2{x}+2cos(x)-1\\\\sin(x)+cos(x)-2sin(x)cos(x)=2cos(x)-2\sin^2{x}\qquad\text{use $1=s^2+c^2$}\\\\sin(x)+2\sin^2{x}-cos(x)-2sin(x)cos(x)=0\\\\sin(x)(1+2sin(x))-cos(x)(1+2sin(x))=0\\\\(sin(x)-cos(x))(1+2sin(x))=0`

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