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

The general solution for 1-(cos^2(x))/(1+sin(x))=2cos(x) is x=1.10714…+pin

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

Solution

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

Solution

x = 1.10714 \dots + πn

Degrees

x = 63.43494 \dots^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

2 cos (x)

from both sides

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

Simplify

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

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

Convert element to fraction:

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

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

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

1 \cdot (1 + sin (x))

Multiply:

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

= (1 + sin (x))

Remove parentheses:

(a) = a

= 1 + sin (x)

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Factor

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

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

Apply exponent rule:

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

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

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

Factor out common term

sin (x)

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

Factor out common term

(1 + sin (x))

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

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

Solving each part separately

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

1 + sin (x) = 0 : x = \frac{3 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

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{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

sin (x) - 2 cos (x) = 0 : x = arctan (2) + πn

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

Rewrite using trig identities

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

Divide both sides by

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

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

Simplify

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

Use the basic trigonometric identity:

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

tan (x) - 2 = 0

tan (x) - 2 = 0

Move

2

to the right side

tan (x) - 2 = 0

Add

2

to both sides

tan (x) - 2 + 2 = 0 + 2

Simplify

tan (x) = 2

tan (x) = 2

Apply trig inverse properties

tan (x) = 2

General solutions for

tan (x) = 2

tan (x) = a \Rightarrow x = arctan (a) + πn

x = arctan (2) + πn

x = arctan (2) + πn

Combine all the solutions

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

Since the equation is undefined for:

\frac{3 π}{2} + 2 πn

x = arctan (2) + πn

Show solutions in decimal form

x = 1.10714 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 1-(cos^2(x))/(1+sin(x))=2cos(x) ?
The general solution for 1-(cos^2(x))/(1+sin(x))=2cos(x) is x=1.10714…+pin