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

The general solution for (1-sin(x))(((1+sin(x)))/(2cos(x)))=1 is No Solution for x\in\mathbb{R}

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

Solution

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

Solution

No Solution for x \in R

Solution steps

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

Subtract

1

from both sides

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

Simplify

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

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

Convert element to fraction:

1 = \frac{1 \cdot 2 cos ( x )}{2 cos ( x )}

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

Since the denominators are equal, combine the fractions:

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

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

Multiply the numbers:

1 \cdot 2 = 2

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

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

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

(1 + sin (x)) (1 - sin (x)) - 2 cos (x) = 0

Rewrite using trig identities

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

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

1 - sin (x)

Multiply by

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

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

Apply Difference of Two Squares Formula:

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

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

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

Use the Pythagorean identity:

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

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

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

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

1 + sin (x)

= cos^{2} (x)

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

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

Solve by substitution

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

Let:

cos (x) = u

u^{2} - 2 u = 0

u^{2} - 2 u = 0 : u = 2, u = 0

u^{2} - 2 u = 0

Solve with the quadratic formula

u^{2} - 2 u = 0

Quadratic Equation Formula:

For

a = 1, b = - 2, c = 0

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot 0 = 2

(- 2)^{2} - 4 \cdot 1 \cdot 0

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

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

= 2^{2} - 4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 2^{2} - 0

2^{2} - 0 = 2^{2}

= 2^{2}

Apply radical rule:

assuming

a \geq 0

= 2

u_{1, 2} = \frac{- ( - 2 ) \pm 2}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 2}{2 \cdot 1}, u_{2} = \frac{- ( - 2 ) - 2}{2 \cdot 1}

u = \frac{- ( - 2 ) + 2}{2 \cdot 1} : 2

\frac{- ( - 2 ) + 2}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{2 + 2}{2 \cdot 1}

Add the numbers:

2 + 2 = 4

= \frac{4}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2

u = \frac{- ( - 2 ) - 2}{2 \cdot 1} : 0

\frac{- ( - 2 ) - 2}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{2 - 2}{2 \cdot 1}

Subtract the numbers:

2 - 2 = 0

= \frac{0}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

The solutions to the quadratic equation are:

u = 2, u = 0

Substitute back

u = cos (x)

cos (x) = 2, cos (x) = 0

cos (x) = 2, cos (x) = 0

cos (x) = 2 :

No Solution

cos (x) = 2

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

General solutions for

cos (x) = 0

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Since the equation is undefined for:

\frac{π}{2} + 2 πn, \frac{3 π}{2} + 2 πn

No Solution for x \in R

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Frequently Asked Questions (FAQ)

What is the general solution for (1-sin(x))(((1+sin(x)))/(2cos(x)))=1 ?
The general solution for (1-sin(x))(((1+sin(x)))/(2cos(x)))=1 is No Solution for x\in\mathbb{R}