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

The general solution for cos^2(x)-sin^2(x)=sec^2(x)-csc^2(x) is x=(3pi)/4+pin,x= pi/4+pin

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cos^2(x)-sin^2(x)=sec^2(x)-csc^2(x)

Solution

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

Solution

x = \frac{3 π}{4} + πn, x = \frac{π}{4} + πn

Degrees

x = 13 5^{\circ} + 18 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

sec^{2} (x) - csc^{2} (x)

from both sides

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

cos^{2} (x) + (\frac{1}{sin ( x )})^{2} - (\frac{1}{cos ( x )})^{2} - sin^{2} (x) : \frac{cos ^{4} ( x ) sin ^{2} ( x ) + cos ^{2} ( x ) - sin ^{2} ( x ) - sin ^{4} ( x ) cos ^{2} ( x )}{sin ^{2} ( x ) cos ^{2} ( x )}

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

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

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Convert element to fraction:

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

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

Least Common Multiplier of

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

1, sin^{2} (x), cos^{2} (x), 1

Lowest Common Multiplier (LCM)

Least Common Multiplier of

1, 1 : 1

1, 1

Least Common Multiplier (LCM)

Prime factorization of

1

Prime factorization of

1

Multiply each factor the greatest number of times it occurs in either

1

1

= 1

Multiply the numbers:

1 = 1

= 1

Compute an expression comprised of factors that appear in at least one of the factored expressions

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

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

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

For

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

multiply the denominator and numerator by

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

\frac{cos ^{2} ( x )}{1} = \frac{cos ^{2} ( x ) sin ^{2} ( x ) cos ^{2} ( x )}{1 \cdot sin ^{2} ( x ) cos ^{2} ( x )} = \frac{cos ^{4} ( x ) sin ^{2} ( x )}{sin ^{2} ( x ) cos ^{2} ( x )}

For

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

multiply the denominator and numerator by

cos^{2} (x)

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

For

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

multiply the denominator and numerator by

sin^{2} (x)

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

For

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

multiply the denominator and numerator by

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

\frac{sin ^{2} ( x )}{1} = \frac{sin ^{2} ( x ) sin ^{2} ( x ) cos ^{2} ( x )}{1 \cdot sin ^{2} ( x ) cos ^{2} ( x )} = \frac{sin ^{4} ( x ) cos ^{2} ( x )}{sin ^{2} ( x ) cos ^{2} ( x )}

= \frac{cos ^{4} ( x ) sin ^{2} ( x )}{sin ^{2} ( x ) cos ^{2} ( x )} + \frac{cos ^{2} ( x )}{sin ^{2} ( x ) cos ^{2} ( x )} - \frac{sin ^{2} ( x )}{sin ^{2} ( x ) cos ^{2} ( x )} - \frac{sin ^{4} ( x ) cos ^{2} ( x )}{sin ^{2} ( 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{cos ^{4} ( x ) sin ^{2} ( x ) + cos ^{2} ( x ) - sin ^{2} ( x ) - sin ^{4} ( x ) cos ^{2} ( x )}{sin ^{2} ( x ) cos ^{2} ( x )}

= \frac{cos ^{4} ( x ) sin ^{2} ( x ) + cos ^{2} ( x ) - sin ^{2} ( x ) - sin ^{4} ( x ) cos ^{2} ( x )}{sin ^{2} ( x ) cos ^{2} ( x )}

\frac{cos ^{2} ( x ) - sin ^{2} ( x ) - cos ^{2} ( x ) sin ^{4} ( x ) + cos ^{4} ( x ) sin ^{2} ( x )}{cos ^{2} ( x ) sin ^{2} ( x )} = 0

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

cos^{2} (x) - sin^{2} (x) - cos^{2} (x) sin^{4} (x) + cos^{4} (x) sin^{2} (x) = 0

Factor

cos^{2} (x) - sin^{2} (x) - cos^{2} (x) sin^{4} (x) + cos^{4} (x) sin^{2} (x) : (1 + cos^{2} (x) sin^{2} (x)) (cos (x) + sin (x)) (cos (x) - sin (x))

cos^{2} (x) - sin^{2} (x) - cos^{2} (x) sin^{4} (x) + cos^{4} (x) sin^{2} (x)

Factor

cos^{2} (x) + cos^{4} (x) sin^{2} (x) : cos^{2} (x) (1 + cos^{2} (x) sin^{2} (x))

cos^{2} (x) + cos^{4} (x) sin^{2} (x)

Apply exponent rule:

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

cos^{4} (x) = cos^{2} (x) cos^{2} (x)

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

Factor out common term

cos^{2} (x)

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

Factor

- sin^{2} (x) - cos^{2} (x) sin^{4} (x) : - sin^{2} (x) (1 + cos^{2} (x) sin^{2} (x))

- sin^{2} (x) - cos^{2} (x) sin^{4} (x)

Apply exponent rule:

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

sin^{4} (x) = sin^{2} (x) sin^{2} (x)

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

Factor out common term

sin^{2} (x)

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

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

Factor out common term

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

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

Factor

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

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

Apply Difference of Two Squares Formula:

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

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

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

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

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

Solving each part separately

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

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

No Solution

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Solve by substitution

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

Let:

sin (x) = u

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

1 + (1 - u^{2}) u^{2} = 0 : u = i \frac{- 1 + 5}{2}, u = - i \frac{- 1 + 5}{2}, u = \frac{1 + 5}{2}, u = - \frac{1 + 5}{2}

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

Expand

1 + (1 - u^{2}) u^{2} : 1 + u^{2} - u^{4}

1 + (1 - u^{2}) u^{2}

= 1 + u^{2} (1 - u^{2})

Expand

u^{2} (1 - u^{2}) : u^{2} - u^{4}

u^{2} (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = u^{2}, b = 1, c = u^{2}

= u^{2} \cdot 1 - u^{2} u^{2}

= 1 \cdot u^{2} - u^{2} u^{2}

Simplify

1 \cdot u^{2} - u^{2} u^{2} : u^{2} - u^{4}

1 \cdot u^{2} - u^{2} u^{2}

1 \cdot u^{2} = u^{2}

1 \cdot u^{2}

Multiply:

1 \cdot u^{2} = u^{2}

= u^{2}

u^{2} u^{2} = u^{4}

u^{2} u^{2}

Apply exponent rule:

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

u^{2} u^{2} = u^{2 + 2}

= u^{2 + 2}

Add the numbers:

2 + 2 = 4

= u^{4}

= u^{2} - u^{4}

= u^{2} - u^{4}

= 1 + u^{2} - u^{4}

1 + u^{2} - u^{4} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

- u^{4} + u^{2} + 1 = 0

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

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

Solve

- v^{2} + v + 1 = 0 : v = - \frac{- 1 + 5}{2}, v = \frac{1 + 5}{2}

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 1, b = 1, c = 1

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

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

1^{2} - 4 (- 1) \cdot 1 = 5

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) \cdot 1

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

Add the numbers:

1 + 4 = 5

= 5

v_{1, 2} = \frac{- 1 \pm 5}{2 ( - 1 )}

Separate the solutions

v_{1} = \frac{- 1 + 5}{2 ( - 1 )}, v_{2} = \frac{- 1 - 5}{2 ( - 1 )}

v = \frac{- 1 + 5}{2 ( - 1 )} : - \frac{- 1 + 5}{2}

\frac{- 1 + 5}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 5}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 + 5}{- 2}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{- 1 + 5}{2}

v = \frac{- 1 - 5}{2 ( - 1 )} : \frac{1 + 5}{2}

\frac{- 1 - 5}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 5}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 - 5}{- 2}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

- 1 - 5 = - (1 + 5)

= \frac{1 + 5}{2}

The solutions to the quadratic equation are:

v = - \frac{- 1 + 5}{2}, v = \frac{1 + 5}{2}

v = - \frac{- 1 + 5}{2}, v = \frac{1 + 5}{2}

Substitute back

v = u^{2},

solve for

u

Solve

u^{2} = - \frac{- 1 + 5}{2} : u = i \frac{- 1 + 5}{2}, u = - i \frac{- 1 + 5}{2}

u^{2} = - \frac{- 1 + 5}{2}

For

x^{2} = f (a)

the solutions are

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

u = - \frac{- 1 + 5}{2}, u = - - \frac{- 1 + 5}{2}

Simplify

- \frac{- 1 + 5}{2} : i \frac{- 1 + 5}{2}

- \frac{- 1 + 5}{2}

Apply radical rule:

- a = - 1 a

- \frac{5 - 1}{2} = - 1 \frac{5 - 1}{2}

= - 1 \frac{5 - 1}{2}

Apply imaginary number rule:

- 1 = i

= i \frac{5 - 1}{2}

Simplify

- - \frac{- 1 + 5}{2} : - i \frac{- 1 + 5}{2}

- - \frac{- 1 + 5}{2}

Simplify

- \frac{- 1 + 5}{2} : i \frac{5 - 1}{2}

- \frac{- 1 + 5}{2}

Apply radical rule:

- a = - 1 a

- \frac{5 - 1}{2} = - 1 \frac{5 - 1}{2}

= - 1 \frac{5 - 1}{2}

Apply imaginary number rule:

- 1 = i

= i \frac{5 - 1}{2}

= - i \frac{5 - 1}{2}

= - i \frac{- 1 + 5}{2}

u = i \frac{- 1 + 5}{2}, u = - i \frac{- 1 + 5}{2}

Solve

u^{2} = \frac{1 + 5}{2} : u = \frac{1 + 5}{2}, u = - \frac{1 + 5}{2}

u^{2} = \frac{1 + 5}{2}

For

x^{2} = f (a)

the solutions are

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

u = \frac{1 + 5}{2}, u = - \frac{1 + 5}{2}

The solutions are

u = i \frac{- 1 + 5}{2}, u = - i \frac{- 1 + 5}{2}, u = \frac{1 + 5}{2}, u = - \frac{1 + 5}{2}

Substitute back

u = sin (x)

sin (x) = i \frac{- 1 + 5}{2}, sin (x) = - i \frac{- 1 + 5}{2}, sin (x) = \frac{1 + 5}{2}, sin (x) = - \frac{1 + 5}{2}

sin (x) = i \frac{- 1 + 5}{2}, sin (x) = - i \frac{- 1 + 5}{2}, sin (x) = \frac{1 + 5}{2}, sin (x) = - \frac{1 + 5}{2}

sin (x) = i \frac{- 1 + 5}{2} :

No Solution

sin (x) = i \frac{- 1 + 5}{2}

No Solution

sin (x) = - i \frac{- 1 + 5}{2} :

No Solution

sin (x) = - i \frac{- 1 + 5}{2}

No Solution

sin (x) = \frac{1 + 5}{2} :

No Solution

sin (x) = \frac{1 + 5}{2}

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = - \frac{1 + 5}{2} :

No Solution

sin (x) = - \frac{1 + 5}{2}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

No Solution

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

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{3 π}{4} + πn

x = \frac{3 π}{4} + 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{3 π}{4} + πn, x = \frac{π}{4} + πn

Popular Examples

2-3sin(x)=0

2 - 3 sin (x) = 0

solvefor y=cos(x),x solve for

y = cos (x), x

(cos(θ))/(sin(θ))=0

\frac{cos ( θ )}{sin ( θ )} = 0

sin(2x)=sin(x),0<= x<2pi

sin (2 x) = sin (x), 0 \leq x < 2 π

tan(θ)= 5/(5sqrt(3))

tan (θ) = \frac{5}{5 3}

Frequently Asked Questions (FAQ)

What is the general solution for cos^2(x)-sin^2(x)=sec^2(x)-csc^2(x) ?
The general solution for cos^2(x)-sin^2(x)=sec^2(x)-csc^2(x) is x=(3pi)/4+pin,x= pi/4+pin