What is the general solution for 2cos^2(θ)-1=sec(θ) ?

The general solution for 2cos^2(θ)-1=sec(θ) is θ=2pin

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Popular Trigonometry >

2cos^2(θ)-1=sec(θ)

Solution

2 cos^{2} (θ) - 1 = sec (θ)

Solution

θ = 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n

Solution steps

2 cos^{2} (θ) - 1 = sec (θ)

Subtract

sec (θ)

from both sides

2 cos^{2} (θ) - 1 - sec (θ) = 0

Rewrite using trig identities

- 1 - sec (θ) + 2 cos^{2} (θ)

Use the basic trigonometric identity:

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

= - 1 - sec (θ) + 2 (\frac{1}{sec ( θ )})^{2}

2 (\frac{1}{sec ( θ )})^{2} = \frac{2}{sec ^{2} ( θ )}

2 (\frac{1}{sec ( θ )})^{2}

(\frac{1}{sec ( θ )})^{2} = \frac{1}{sec ^{2} ( θ )}

(\frac{1}{sec ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{sec ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sec ^{2} ( θ )}

= 2 \cdot \frac{1}{sec ^{2} ( θ )}

Multiply fractions:

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

= \frac{1 \cdot 2}{sec ^{2} ( θ )}

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2}{sec ^{2} ( θ )}

= - 1 - sec (θ) + \frac{2}{sec ^{2} ( θ )}

- 1 + \frac{2}{sec ^{2} ( θ )} - sec (θ) = 0

Solve by substitution

- 1 + \frac{2}{sec ^{2} ( θ )} - sec (θ) = 0

Let:

sec (θ) = u

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

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

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

Multiply both sides by

u^{2}

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

Multiply both sides by

u^{2}

- 1 \cdot u^{2} + \frac{2}{u ^{2}} u^{2} - u u^{2} = 0 \cdot u^{2}

Simplify

- 1 \cdot u^{2} + \frac{2}{u ^{2}} u^{2} - u u^{2} = 0 \cdot u^{2}

Simplify

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

- 1 \cdot u^{2}

Multiply:

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

= - u^{2}

Simplify

\frac{2}{u ^{2}} u^{2} : 2

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

Multiply fractions:

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

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

Cancel the common factor:

u^{2}

= 2

Simplify

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

- u 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}

Simplify

0 \cdot u^{2} : 0

0 \cdot u^{2}

Apply rule

0 \cdot a = 0

= 0

- u^{2} + 2 - u^{3} = 0

- u^{2} + 2 - u^{3} = 0

- u^{2} + 2 - u^{3} = 0

Solve

- u^{2} + 2 - u^{3} = 0 : u = 1, u = - 1 + i, u = - 1 - i

- u^{2} + 2 - u^{3} = 0

Write in the standard form

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

- u^{3} - u^{2} + 2 = 0

Factor

- u^{3} - u^{2} + 2 : - (u - 1) (u^{2} + 2 u + 2)

- u^{3} - u^{2} + 2

Factor out common term

- 1

= - (u^{3} + u^{2} - 2)

Factor

u^{3} + u^{2} - 2 : (u - 1) (u^{2} + 2 u + 2)

u^{3} + u^{2} - 2

Use the rational root theorem

a_{0} = 2, a_{n} = 1

The dividers of

a_{0} : 1, 2,

The dividers of

a_{n} : 1

Therefore, check the following rational numbers:

\pm \frac{1 , 2}{1}

\frac{1}{1}

is a root of the expression, so factor out

u - 1

= (u - 1) \frac{u ^{3} + u ^{2} - 2}{u - 1}

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

\frac{u ^{3} + u ^{2} - 2}{u - 1}

Divide

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

Divide the leading coefficients of the numerator

u^{3} + u^{2} - 2

and the divisor

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

Quotient = u^{2}

Multiply

u - 1

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

Subtract

u^{3} - u^{2}

from

u^{3} + u^{2} - 2

to get new remainder

Remainder = 2 u^{2} - 2

Therefore

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

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

Divide

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

Divide the leading coefficients of the numerator

2 u^{2} - 2

and the divisor

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

Quotient = 2 u

Multiply

u - 1

2 u : 2 u^{2} - 2 u

Subtract

2 u^{2} - 2 u

from

2 u^{2} - 2

to get new remainder

Remainder = 2 u - 2

Therefore

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

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

Divide

\frac{2 u - 2}{u - 1} : \frac{2 u - 2}{u - 1} = 2

Divide the leading coefficients of the numerator

2 u - 2

and the divisor

u - 1 : \frac{2 u}{u} = 2

Quotient = 2

Multiply

u - 1

2 : 2 u - 2

Subtract

2 u - 2

from

2 u - 2

to get new remainder

Remainder = 0

Therefore

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

= u^{2} + 2 u + 2

= u^{2} + 2 u + 2

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

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

u - 1 = 0 : u = 1

u - 1 = 0

Move

1

to the right side

u - 1 = 0

Add

1

to both sides

u - 1 + 1 = 0 + 1

Simplify

u = 1

u = 1

Solve

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 1, b = 2, c = 2

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot 2}{2 \cdot 1}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot 2}{2 \cdot 1}

Simplify

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

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

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 2^{2} - 8

Apply imaginary number rule:

- a = i a

= i 8 - 2^{2}

- 2^{2} + 8 = 2

- 2^{2} + 8

2^{2} = 4

= - 4 + 8

Add/Subtract the numbers:

- 4 + 8 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

n a^{n} = a

2^{2} = 2

= 2

= 2 i

u_{1, 2} = \frac{- 2 \pm 2 i}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 2 + 2 i}{2 \cdot 1}, u_{2} = \frac{- 2 - 2 i}{2 \cdot 1}

u = \frac{- 2 + 2 i}{2 \cdot 1} : - 1 + i

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

Multiply the numbers:

2 \cdot 1 = 2

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

Factor

- 2 + 2 i : 2 (- 1 + i)

- 2 + 2 i

Rewrite as

= - 2 \cdot 1 + 2 i

Factor out common term

2

= 2 (- 1 + i)

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

Divide the numbers:

\frac{2}{2} = 1

= - 1 + i

u = \frac{- 2 - 2 i}{2 \cdot 1} : - 1 - i

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

Multiply the numbers:

2 \cdot 1 = 2

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

Factor

- 2 - 2 i : - 2 (1 + i)

- 2 - 2 i

Rewrite as

= - 2 \cdot 1 - 2 i

Factor out common term

2

= - 2 (1 + i)

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

Divide the numbers:

\frac{2}{2} = 1

= - (1 + i)

Negate

- (1 + i) = - 1 - i

= - 1 - i

The solutions to the quadratic equation are:

u = - 1 + i, u = - 1 - i

The solutions are

u = 1, u = - 1 + i, u = - 1 - i

u = 1, u = - 1 + i, u = - 1 - i

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

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

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 = - 1 + i, u = - 1 - i

Substitute back

u = sec (θ)

sec (θ) = 1, sec (θ) = - 1 + i, sec (θ) = - 1 - i

sec (θ) = 1, sec (θ) = - 1 + i, sec (θ) = - 1 - i

sec (θ) = 1 : θ = 2 πn

sec (θ) = 1

General solutions for

sec (θ) = 1

sec (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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} 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} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

θ = 0 + 2 πn

θ = 0 + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

sec (θ) = - 1 + i :

No Solution

sec (θ) = - 1 + i

No Solution

sec (θ) = - 1 - i :

No Solution

sec (θ) = - 1 - i

No Solution

Combine all the solutions

θ = 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(θ)-1=sec(θ) ?
The general solution for 2cos^2(θ)-1=sec(θ) is θ=2pin