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

The general solution for cos(x)cos(x)=((1-sin^2(x)))/(sin(x)) is x= pi/2+2pin,x=(3pi)/2+2pin

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

Solution

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

Solution

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

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

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

from both sides

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Factor

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

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

Factor out common term

cos^{2} (x)

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

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

Solving each part separately

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

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

cos^{2} (x) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

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

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

sin (x) - 1 = 0

Move

1

to the right side

sin (x) - 1 = 0

Add

1

to both sides

sin (x) - 1 + 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{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

Combine all the solutions

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

Popular Examples

r^2=a^2(sin(x)+cos(x))

r^{2} = a^{2} (sin (x) + cos (x))

sec(x)-2=0,0<= x<= 2pi

sec (x) - 2 = 0, 0 \leq x \leq 2 π

(cos(t)-3)(2sin^2(t)-1)=0

(cos (t) - 3) (2 sin^{2} (t) - 1) = 0

cot(x)=3.2404

cot (x) = 3.2404

cot(2θ)= 5/12

cot (2 θ) = \frac{5}{12}

Frequently Asked Questions (FAQ)

What is the general solution for cos(x)cos(x)=((1-sin^2(x)))/(sin(x)) ?
The general solution for cos(x)cos(x)=((1-sin^2(x)))/(sin(x)) is x= pi/2+2pin,x=(3pi)/2+2pin