Question

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

Answer 1

Answer:

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

Step-by-step explanation:

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

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