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

The general solution for (1-sin(x))(1+cos(x))=cos^2(x) is x= pi/2+2pin,x= pi/4+pin

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

Solution

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

Solution

x = \frac{π}{2} + 2 πn, x = \frac{π}{4} + πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

cos^{2} (x)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

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

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

Expand

(1 + cos (x)) (1 - sin (x)) : 1 - sin (x) + cos (x) - cos (x) sin (x)

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

Apply FOIL method:

(a + b) (c + d) = a c + a d + b c + b d

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

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

Apply minus-plus rules

+ (- a) = - a

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

Simplify

1 \cdot 1 - 1 \cdot sin (x) + 1 \cdot cos (x) - cos (x) sin (x) : 1 - sin (x) + cos (x) - cos (x) sin (x)

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

Multiply the numbers:

1 \cdot 1 = 1

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

Multiply:

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

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

Multiply:

1 \cdot cos (x) = cos (x)

= 1 - sin (x) + cos (x) - cos (x) sin (x)

= 1 - sin (x) + cos (x) - cos (x) sin (x)

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

Simplify

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

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

Group like terms

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

- 1 + 1 = 0

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

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

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

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

Factor

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

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

Factor out common term

cos (x)

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

Apply exponent rule:

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

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

= cos (x) (1 - sin (x)) - sin (x) + sin (x) sin (x)

Factor out common term

sin (x)

= cos (x) (1 - sin (x)) + sin (x) (- 1 + sin (x))

Rewrite as

= (1 - sin (x)) cos (x) - (1 - sin (x)) sin (x)

Factor out common term

(1 - sin (x))

= (1 - sin (x)) (cos (x) - sin (x))

(1 - sin (x)) (cos (x) - sin (x)) = 0

Solving each part separately

1 - sin (x) = 0 or cos (x) - sin (x) = 0

1 - sin (x) = 0 : x = \frac{π}{2} + 2 πn

1 - sin (x) = 0

Move

1

to the right side

1 - sin (x) = 0

Subtract

1

from both sides

1 - sin (x) - 1 = 0 - 1

Simplify

- sin (x) = - 1

- sin (x) = - 1

Divide both sides by

- 1

- sin (x) = - 1

Divide both sides by

- 1

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

Simplify

sin (x) = 1

sin (x) = 1

General solutions for

sin (x) = 1

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 = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

cos (x) - sin (x) = 0 : x = \frac{π}{4} + πn

cos (x) - sin (x) = 0

Rewrite using trig identities

cos (x) - sin (x) = 0

Divide both sides by

cos (x), cos (x) \neq = 0

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

Simplify

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

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

1 - tan (x) = 0

1 - tan (x) = 0

Move

1

to the right side

1 - tan (x) = 0

Subtract

1

from both sides

1 - tan (x) - 1 = 0 - 1

Simplify

- tan (x) = - 1

- tan (x) = - 1

Divide both sides by

- 1

- tan (x) = - 1

Divide both sides by

- 1

\frac{- tan ( x )}{- 1} = \frac{- 1}{- 1}

Simplify

tan (x) = 1

tan (x) = 1

General solutions for

tan (x) = 1

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{π}{4} + πn

Popular Examples

sin(θ)=(-1)/(sqrt(2)),0<= θ<= 4pi

sin (θ) = \frac{- 1}{2}, 0 \leq θ \leq 4 π

0=2cos(θ)

0 = 2 cos (θ)

-2sin(x)-2cos(2x)=0

- 2 sin (x) - 2 cos (2 x) = 0

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

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

solvefor y,2cos(5x+y)=0 solve for

y, 2 cos (5 x + y) = 0

Frequently Asked Questions (FAQ)

What is the general solution for (1-sin(x))(1+cos(x))=cos^2(x) ?
The general solution for (1-sin(x))(1+cos(x))=cos^2(x) is x= pi/2+2pin,x= pi/4+pin