What is the general solution for (1-cos^2(x))/(1+sin(x))=sin(x) ?

The general solution for (1-cos^2(x))/(1+sin(x))=sin(x) is x=2pin,x=pi+2pin

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(1-cos^2(x))/(1+sin(x))=sin(x)

Solution

\frac{1 - cos ^{2} ( x )}{1 + sin ( x )} = sin (x)

Solution

x = 2 πn, x = π + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

\frac{1 - cos ^{2} ( x )}{1 + sin ( x )} = sin (x)

Subtract

sin (x)

from both sides

\frac{1 - cos ^{2} ( x )}{1 + sin ( x )} - sin (x) = 0

Simplify

\frac{1 - cos ^{2} ( x )}{1 + sin ( x )} - sin (x) : \frac{1 - cos ^{2} ( x ) - sin ( x ) ( 1 + sin ( x ) )}{1 + sin ( x )}

\frac{1 - cos ^{2} ( x )}{1 + sin ( x )} - sin (x)

Convert element to fraction:

sin (x) = \frac{sin ( x ) ( 1 + sin ( x ) )}{1 + sin ( x )}

= \frac{1 - cos ^{2} ( x )}{1 + sin ( x )} - \frac{sin ( x ) ( 1 + sin ( x ) )}{1 + sin ( x )}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 - cos ^{2} ( x ) - sin ( x ) ( 1 + sin ( x ) )}{1 + sin ( x )}

\frac{1 - cos ^{2} ( x ) - sin ( x ) ( 1 + sin ( x ) )}{1 + sin ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

1 - cos^{2} (x) - sin (x) (1 + sin (x)) = 0

Rewrite using trig identities

1 - cos^{2} (x) - (1 + sin (x)) sin (x)

Use the Pythagorean identity:

1 = cos^{2} (x) + sin^{2} (x)

1 - cos^{2} (x) = sin^{2} (x)

= - sin (x) (1 + sin (x)) + sin^{2} (x)

Simplify

- sin (x) (1 + sin (x)) + sin^{2} (x) : - sin (x)

- sin (x) (1 + sin (x)) + sin^{2} (x)

Expand

- sin (x) (1 + sin (x)) : - sin (x) - sin^{2} (x)

- sin (x) (1 + sin (x))

Apply the distributive law:

a (b + c) = ab + a c

a = - sin (x), b = 1, c = sin (x)

= - sin (x) \cdot 1 + (- sin (x)) sin (x)

Apply minus-plus rules

+ (- a) = - a

= - 1 \cdot sin (x) - sin (x) sin (x)

Simplify

- 1 \cdot sin (x) - sin (x) sin (x) : - sin (x) - sin^{2} (x)

- 1 \cdot sin (x) - sin (x) sin (x)

1 \cdot sin (x) = sin (x)

1 \cdot sin (x)

Multiply:

1 \cdot sin (x) = sin (x)

= sin (x)

sin (x) sin (x) = sin^{2} (x)

sin (x) sin (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

sin (x) sin (x) = sin^{1 + 1} (x)

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

= - sin (x) - sin^{2} (x)

= - sin (x) - sin^{2} (x)

= - sin (x) - sin^{2} (x) + sin^{2} (x)

Add similar elements:

- sin^{2} (x) + sin^{2} (x) = 0

= - sin (x)

= - sin (x)

- sin (x) = 0

Divide both sides by

- 1

- sin (x) = 0

Divide both sides by

- 1

\frac{- sin ( x )}{- 1} = \frac{0}{- 1}

Simplify

sin (x) = 0

sin (x) = 0

General solutions for

sin (x) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

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cos (x^{2}) = - \frac{2}{2}

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4sin(θ)+4=0

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sin^2(a)+2cos(a)= 49/25

sin^{2} (a) + 2 cos (a) = \frac{49}{25}

Frequently Asked Questions (FAQ)

What is the general solution for (1-cos^2(x))/(1+sin(x))=sin(x) ?
The general solution for (1-cos^2(x))/(1+sin(x))=sin(x) is x=2pin,x=pi+2pin