What is the general solution for sec^2(x)tan^2(x)=sin^2(x) ?

The general solution for sec^2(x)tan^2(x)=sin^2(x) is x=2pin,x=pi+2pin

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sec^2(x)tan^2(x)=sin^2(x)

Solution

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

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

Subtract

sin^{2} (x)

from both sides

sec^{2} (x) tan^{2} (x) - sin^{2} (x) = 0

Factor

sec^{2} (x) tan^{2} (x) - sin^{2} (x) : (sec (x) tan (x) + sin (x)) (sec (x) tan (x) - sin (x))

sec^{2} (x) tan^{2} (x) - sin^{2} (x)

Rewrite

sec^{2} (x) tan^{2} (x)

(sec (x) tan (x))^{2}

sec^{2} (x) tan^{2} (x)

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

sec^{2} (x) tan^{2} (x) = (sec (x) tan (x))^{2}

= (sec (x) tan (x))^{2}

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

Apply Difference of Two Squares Formula:

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

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

= (sec (x) tan (x) + sin (x)) (sec (x) tan (x) - sin (x))

(sec (x) tan (x) + sin (x)) (sec (x) tan (x) - sin (x)) = 0

Solving each part separately

sec (x) tan (x) + sin (x) = 0 or sec (x) tan (x) - sin (x) = 0

sec (x) tan (x) + sin (x) = 0 : x = 2 πn, x = π + 2 πn

sec (x) tan (x) + sin (x) = 0

Express with sin, cos

sin (x) + sec (x) tan (x)

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

\frac{1}{cos ( x )} \cdot \frac{sin ( x )}{cos ( x )}

Multiply fractions:

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

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

Multiply:

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

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

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

cos (x) cos (x)

Apply exponent rule:

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

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

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

Factor

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

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

Factor out common term

sin (x)

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

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

Solving each part separately

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

sin (x) = 0 : x = 2 πn, x = π + 2 πn

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

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

No Solution

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

Solve by substitution

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

Let:

cos (x) = u

u^{2} + 1 = 0

u^{2} + 1 = 0 : u = i, u = - i

u^{2} + 1 = 0

Move

1

to the right side

u^{2} + 1 = 0

Subtract

1

from both sides

u^{2} + 1 - 1 = 0 - 1

Simplify

u^{2} = - 1

u^{2} = - 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = - 1, u = - - 1

Simplify

- 1 : i

- 1

Apply imaginary number rule:

- 1 = i

= i

Simplify

- - 1 : - i

- - 1

Apply imaginary number rule:

- 1 = i

= - i

u = i, u = - i

Substitute back

u = cos (x)

cos (x) = i, cos (x) = - i

cos (x) = i, cos (x) = - i

cos (x) = i :

No Solution

cos (x) = i

No Solution

cos (x) = - i :

No Solution

cos (x) = - i

No Solution

Combine all the solutions

No Solution

Combine all the solutions

x = 2 πn, x = π + 2 πn

sec (x) tan (x) - sin (x) = 0 : x = 2 πn, x = π + 2 πn

sec (x) tan (x) - sin (x) = 0

Express with sin, cos

- sin (x) + sec (x) tan (x)

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

\frac{1}{cos ( x )} \cdot \frac{sin ( x )}{cos ( x )}

Multiply fractions:

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

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

Multiply:

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

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

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

cos (x) cos (x)

Apply exponent rule:

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

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

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

Factor

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

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

Factor out common term

- sin (x)

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

Factor

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

cos^{2} (x) - 1

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

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

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

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

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

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

Solving each part separately

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

sin (x) = 0 : x = 2 πn, x = π + 2 πn

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

cos (x) + 1 = 0 : x = π + 2 πn

cos (x) + 1 = 0

Move

1

to the right side

cos (x) + 1 = 0

Subtract

1

from both sides

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

Simplify

cos (x) = - 1

cos (x) = - 1

General solutions for

cos (x) = - 1

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 = π + 2 πn

x = π + 2 πn

cos (x) - 1 = 0 : x = 2 πn

cos (x) - 1 = 0

Move

1

to the right side

cos (x) - 1 = 0

Add

1

to both sides

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

Simplify

cos (x) = 1

cos (x) = 1

General solutions for

cos (x) = 1

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 = 0 + 2 πn

x = 0 + 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

Combine all the solutions

x = 2 πn, x = π + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for sec^2(x)tan^2(x)=sin^2(x) ?
The general solution for sec^2(x)tan^2(x)=sin^2(x) is x=2pin,x=pi+2pin