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

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

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Popular Trigonometry >

1/(2cos^2(x-1))=(1+tan^2(x))/(2sec^2(x))

Solution

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

Solution

x = π + 2 πn + 1, x = 2 πn + 1

Degrees

x = 237.29577 \dots^{\circ} + 36 0^{\circ} n, x = 57.29577 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

\frac{1 + tan ^{2} ( x )}{2 sec ^{2} ( x )}

from both sides

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

Simplify

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

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

Least Common Multiplier of

2 cos^{2} (x - 1), 2 sec^{2} (x) : 2 cos^{2} (x - 1) sec^{2} (x)

2 cos^{2} (x - 1), 2 sec^{2} (x)

Lowest Common Multiplier (LCM)

Least Common Multiplier of

2, 2 : 2

2, 2

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

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

2

2

= 2

Multiply the numbers:

2 = 2

= 2

Compute an expression comprised of factors that appear either in

2 cos^{2} (x - 1)

2 sec^{2} (x)

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

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

For

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

multiply the denominator and numerator by

sec^{2} (x)

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

For

\frac{1 + tan ^{2} ( x )}{2 sec ^{2} ( x )} :

multiply the denominator and numerator by

cos^{2} (x - 1)

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

Rewrite using trig identities

sec^{2} (x) - (1 + tan^{2} (x)) cos^{2} (- 1 + x)

Use the Pythagorean identity:

tan^{2} (x) + 1 = sec^{2} (x)

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

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

Factor

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

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

Factor out common term

- sec^{2} (x)

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

Factor

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

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

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

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

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

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

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

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

Solving each part separately

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

sec^{2} (x) = 0 :

No Solution

sec^{2} (x) = 0

Apply rule

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

sec (x) = 0

sec (x) \leq - 1 or sec (x) \geq 1

No Solution

cos (- 1 + x) + 1 = 0 : x = π + 2 πn + 1

cos (- 1 + x) + 1 = 0

Move

1

to the right side

cos (- 1 + x) + 1 = 0

Subtract

1

from both sides

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

Simplify

cos (- 1 + x) = - 1

cos (- 1 + x) = - 1

General solutions for

cos (- 1 + 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}

- 1 + x = π + 2 πn

- 1 + x = π + 2 πn

Solve

- 1 + x = π + 2 πn : x = π + 2 πn + 1

- 1 + x = π + 2 πn

Move

1

to the right side

- 1 + x = π + 2 πn

Add

1

to both sides

- 1 + x + 1 = π + 2 πn + 1

Simplify

x = π + 2 πn + 1

x = π + 2 πn + 1

x = π + 2 πn + 1

cos (- 1 + x) - 1 = 0 : x = 2 πn + 1

cos (- 1 + x) - 1 = 0

Move

1

to the right side

cos (- 1 + x) - 1 = 0

Add

1

to both sides

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

Simplify

cos (- 1 + x) = 1

cos (- 1 + x) = 1

General solutions for

cos (- 1 + 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}

- 1 + x = 0 + 2 πn

- 1 + x = 0 + 2 πn

Solve

- 1 + x = 0 + 2 πn : x = 2 πn + 1

- 1 + x = 0 + 2 πn

0 + 2 πn = 2 πn

- 1 + x = 2 πn

Move

1

to the right side

- 1 + x = 2 πn

Add

1

to both sides

- 1 + x + 1 = 2 πn + 1

Simplify

x = 2 πn + 1

x = 2 πn + 1

x = 2 πn + 1

Combine all the solutions

x = π + 2 πn + 1, x = 2 πn + 1

Graph

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Frequently Asked Questions (FAQ)

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