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Popular >

cos^4(x)+2cos^2(x)=1

Solution

cos^{4} (x) + 2 cos^{2} (x) = 1

Solution

x = 0.87161 \dots + 2 πn, x = 2 π - 0.87161 \dots + 2 πn, x = 2.26998 \dots + 2 πn, x = - 2.26998 \dots + 2 πn

Degrees

x = 49.93964 \dots^{\circ} + 36 0^{\circ} n, x = 310.06035 \dots^{\circ} + 36 0^{\circ} n, x = 130.06035 \dots^{\circ} + 36 0^{\circ} n, x = - 130.06035 \dots^{\circ} + 36 0^{\circ} n

Solution steps

cos^{4} (x) + 2 cos^{2} (x) = 1

Solve by substitution

cos^{4} (x) + 2 cos^{2} (x) = 1

Let:

cos (x) = u

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

u^{4} + 2 u^{2} = 1 : u = - 1 + 2, u = - - 1 + 2, u = i 1 + 2, u = - i 1 + 2

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

Move

1

to the left side

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

Subtract

1

from both sides

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

Simplify

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

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

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

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

Solve

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 1, b = 2, c = - 1

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

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

2^{2} - 4 \cdot 1 \cdot (- 1) = 22

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

Apply rule

- (- a) = a

= 2^{2} + 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} + 4

2^{2} = 4

= 4 + 4

Add the numbers:

4 + 4 = 8

= 8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

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

= 2^{2} \cdot 2

Apply radical rule:

= 2 2^{2}

Apply radical rule:

2^{2} = 2

= 22

v_{1, 2} = \frac{- 2 \pm 2 2}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- 2 + 2 2}{2 \cdot 1}, v_{2} = \frac{- 2 - 2 2}{2 \cdot 1}

v = \frac{- 2 + 2 2}{2 \cdot 1} : - 1 + 2

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

Multiply the numbers:

2 \cdot 1 = 2

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

Factor

- 2 + 22 : 2 (- 1 + 2)

- 2 + 22

Rewrite as

= - 2 \cdot 1 + 22

Factor out common term

2

= 2 (- 1 + 2)

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

Divide the numbers:

\frac{2}{2} = 1

= - 1 + 2

v = \frac{- 2 - 2 2}{2 \cdot 1} : - 1 - 2

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

Multiply the numbers:

2 \cdot 1 = 2

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

Factor

- 2 - 22 : - 2 (1 + 2)

- 2 - 22

Rewrite as

= - 2 \cdot 1 - 22

Factor out common term

2

= - 2 (1 + 2)

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

Divide the numbers:

\frac{2}{2} = 1

= - (1 + 2)

Negate

- (1 + 2) = - 1 - 2

= - 1 - 2

The solutions to the quadratic equation are:

v = - 1 + 2, v = - 1 - 2

v = - 1 + 2, v = - 1 - 2

Substitute back

v = u^{2},

solve for

u

Solve

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

u^{2} = - 1 + 2

For

x^{2} = f (a)

the solutions are

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

u = - 1 + 2, u = - - 1 + 2

Solve

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

u^{2} = - 1 - 2

For

x^{2} = f (a)

the solutions are

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

u = - 1 - 2, u = - - 1 - 2

Simplify

- 1 - 2 : i 1 + 2

- 1 - 2

Apply imaginary number rule:

- a = i a

= i 1 + 2

Simplify

- - 1 - 2 : - i 1 + 2

- - 1 - 2

Apply imaginary number rule:

- a = i a

= - i 1 + 2

u = i 1 + 2, u = - i 1 + 2

The solutions are

u = - 1 + 2, u = - - 1 + 2, u = i 1 + 2, u = - i 1 + 2

Substitute back

u = cos (x)

cos (x) = - 1 + 2, cos (x) = - - 1 + 2, cos (x) = i 1 + 2, cos (x) = - i 1 + 2

cos (x) = - 1 + 2, cos (x) = - - 1 + 2, cos (x) = i 1 + 2, cos (x) = - i 1 + 2

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

cos (x) = - 1 + 2

Apply trig inverse properties

cos (x) = - 1 + 2

General solutions for

cos (x) = - 1 + 2

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (- 1 + 2) + 2 πn, x = 2 π - arccos (- 1 + 2) + 2 πn

x = arccos (- 1 + 2) + 2 πn, x = 2 π - arccos (- 1 + 2) + 2 πn

cos (x) = - - 1 + 2 : x = arccos (- - 1 + 2) + 2 πn, x = - arccos (- - 1 + 2) + 2 πn

cos (x) = - - 1 + 2

Apply trig inverse properties

cos (x) = - - 1 + 2

General solutions for

cos (x) = - - 1 + 2

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (- - 1 + 2) + 2 πn, x = - arccos (- - 1 + 2) + 2 πn

x = arccos (- - 1 + 2) + 2 πn, x = - arccos (- - 1 + 2) + 2 πn

cos (x) = i 1 + 2 :

No Solution

cos (x) = i 1 + 2

No Solution

cos (x) = - i 1 + 2 :

No Solution

cos (x) = - i 1 + 2

No Solution

Combine all the solutions

x = arccos (- 1 + 2) + 2 πn, x = 2 π - arccos (- 1 + 2) + 2 πn, x = arccos (- - 1 + 2) + 2 πn, x = - arccos (- - 1 + 2) + 2 πn

Show solutions in decimal form

x = 0.87161 \dots + 2 πn, x = 2 π - 0.87161 \dots + 2 πn, x = 2.26998 \dots + 2 πn, x = - 2.26998 \dots + 2 πn

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