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

The general solution for cos(x)= 1/(cos^2(x)) is x=2pin

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

Solution

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

Solution

x = 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

cos (x) = u

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

u = \frac{1}{u ^{2}} : u = 1, u = - \frac{1}{2} + i \frac{3}{2}, u = - \frac{1}{2} - i \frac{3}{2}

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

Multiply both sides by

u^{2}

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

Multiply both sides by

u^{2}

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

Simplify

u u^{2} : u^{3}

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

Apply exponent rule:

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

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

= u^{1 + 2}

Add the numbers:

1 + 2 = 3

= u^{3}

u^{3} = 1

u^{3} = 1

Solve

u^{3} = 1 : u = 1, u = - \frac{1}{2} + i \frac{3}{2}, u = - \frac{1}{2} - i \frac{3}{2}

u^{3} = 1

For

x^{3} = f (a)

the solutions are

u = 1, u = \frac{- 1 + 3 i}{2}, u = \frac{- 1 - 3 i}{2}

Simplify

\frac{- 1 + 3 i}{2} : - \frac{1}{2} + i \frac{3}{2}

\frac{- 1 + 3 i}{2}

Rewrite

\frac{- 1 + 3 i}{2}

in standard complex form:

- \frac{1}{2} + \frac{3}{2} i

\frac{- 1 + 3 i}{2}

Apply the fraction rule:

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

\frac{- 1 + 3 i}{2} = - \frac{1}{2} + \frac{3 i}{2}

= - \frac{1}{2} + \frac{3 i}{2}

= - \frac{1}{2} + \frac{3}{2} i

Simplify

\frac{- 1 - 3 i}{2} : - \frac{1}{2} - i \frac{3}{2}

\frac{- 1 - 3 i}{2}

Rewrite

\frac{- 1 - 3 i}{2}

in standard complex form:

- \frac{1}{2} - \frac{3}{2} i

\frac{- 1 - 3 i}{2}

Apply the fraction rule:

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

\frac{- 1 - 3 i}{2} = - \frac{1}{2} - \frac{3 i}{2}

= - \frac{1}{2} - \frac{3 i}{2}

= - \frac{1}{2} - \frac{3}{2} i

u = 1, u = - \frac{1}{2} + i \frac{3}{2}, u = - \frac{1}{2} - i \frac{3}{2}

u = 1, u = - \frac{1}{2} + i \frac{3}{2}, u = - \frac{1}{2} - i \frac{3}{2}

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

\frac{1}{u ^{2}}

and compare to zero

Solve

u^{2} = 0 : u = 0

u^{2} = 0

Apply rule

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

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = 1, u = - \frac{1}{2} + i \frac{3}{2}, u = - \frac{1}{2} - i \frac{3}{2}

Substitute back

u = cos (x)

cos (x) = 1, cos (x) = - \frac{1}{2} + i \frac{3}{2}, cos (x) = - \frac{1}{2} - i \frac{3}{2}

cos (x) = 1, cos (x) = - \frac{1}{2} + i \frac{3}{2}, cos (x) = - \frac{1}{2} - i \frac{3}{2}

cos (x) = 1 : x = 2 πn

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

cos (x) = - \frac{1}{2} + i \frac{3}{2} :

No Solution

cos (x) = - \frac{1}{2} + i \frac{3}{2}

No Solution

cos (x) = - \frac{1}{2} - i \frac{3}{2} :

No Solution

cos (x) = - \frac{1}{2} - i \frac{3}{2}

No Solution

Combine all the solutions

x = 2 πn

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Frequently Asked Questions (FAQ)

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