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Popular >

(-1-3cos(x))^2+cos^2(x)=1

Solution

(- 1 - 3 cos (x))^{2} + cos^{2} (x) = 1

Solution

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = 2.21429 \dots + 2 πn, x = - 2.21429 \dots + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n, x = 126.86989 \dots^{\circ} + 36 0^{\circ} n, x = - 126.86989 \dots^{\circ} + 36 0^{\circ} n

Solution steps

(- 1 - 3 cos (x))^{2} + cos^{2} (x) = 1

Solve by substitution

(- 1 - 3 cos (x))^{2} + cos^{2} (x) = 1

Let:

cos (x) = u

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

(- 1 - 3 u)^{2} + u^{2} = 1 : u = 0, u = - \frac{3}{5}

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

Expand

(- 1 - 3 u)^{2} + u^{2} : 1 + 6 u + 10 u^{2}

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

(- 1 - 3 u)^{2} : 1 + 6 u + 9 u^{2}

Apply Perfect Square Formula:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = - 1, b = 3 u

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

Simplify

(- 1)^{2} - 2 (- 1) \cdot 3 u + (3 u)^{2} : 1 + 6 u + 9 u^{2}

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

Apply rule

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

2 \cdot 1 \cdot 3 u = 6 u

2 \cdot 1 \cdot 3 u

Multiply the numbers:

2 \cdot 1 \cdot 3 = 6

= 6 u

(3 u)^{2} = 9 u^{2}

(3 u)^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 3^{2} u^{2}

3^{2} = 9

= 9 u^{2}

= 1 + 6 u + 9 u^{2}

= 1 + 6 u + 9 u^{2}

= 1 + 6 u + 9 u^{2} + u^{2}

Add similar elements:

9 u^{2} + u^{2} = 10 u^{2}

= 1 + 6 u + 10 u^{2}

1 + 6 u + 10 u^{2} = 1

Move

1

to the left side

1 + 6 u + 10 u^{2} = 1

Subtract

1

from both sides

1 + 6 u + 10 u^{2} - 1 = 1 - 1

Simplify

10 u^{2} + 6 u = 0

10 u^{2} + 6 u = 0

Solve with the quadratic formula

10 u^{2} + 6 u = 0

Quadratic Equation Formula:

For

a = 10, b = 6, c = 0

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 10 \cdot 0}{2 \cdot 10}

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 10 \cdot 0}{2 \cdot 10}

6^{2} - 4 \cdot 10 \cdot 0 = 6

6^{2} - 4 \cdot 10 \cdot 0

Apply rule

0 \cdot a = 0

= 6^{2} - 0

6^{2} - 0 = 6^{2}

= 6^{2}

Apply radical rule:

assuming

a \geq 0

= 6

u_{1, 2} = \frac{- 6 \pm 6}{2 \cdot 10}

Separate the solutions

u_{1} = \frac{- 6 + 6}{2 \cdot 10}, u_{2} = \frac{- 6 - 6}{2 \cdot 10}

u = \frac{- 6 + 6}{2 \cdot 10} : 0

\frac{- 6 + 6}{2 \cdot 10}

Add/Subtract the numbers:

- 6 + 6 = 0

= \frac{0}{2 \cdot 10}

Multiply the numbers:

2 \cdot 10 = 20

= \frac{0}{20}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

u = \frac{- 6 - 6}{2 \cdot 10} : - \frac{3}{5}

\frac{- 6 - 6}{2 \cdot 10}

Subtract the numbers:

- 6 - 6 = - 12

= \frac{- 12}{2 \cdot 10}

Multiply the numbers:

2 \cdot 10 = 20

= \frac{- 12}{20}

Apply the fraction rule:

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

= - \frac{12}{20}

Cancel the common factor:

4

= - \frac{3}{5}

The solutions to the quadratic equation are:

u = 0, u = - \frac{3}{5}

Substitute back

u = cos (x)

cos (x) = 0, cos (x) = - \frac{3}{5}

cos (x) = 0, cos (x) = - \frac{3}{5}

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

General solutions for

cos (x) = 0

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 = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = - \frac{3}{5} : x = arccos (- \frac{3}{5}) + 2 πn, x = - arccos (- \frac{3}{5}) + 2 πn

cos (x) = - \frac{3}{5}

Apply trig inverse properties

cos (x) = - \frac{3}{5}

General solutions for

cos (x) = - \frac{3}{5}

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

x = arccos (- \frac{3}{5}) + 2 πn, x = - arccos (- \frac{3}{5}) + 2 πn

x = arccos (- \frac{3}{5}) + 2 πn, x = - arccos (- \frac{3}{5}) + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = arccos (- \frac{3}{5}) + 2 πn, x = - arccos (- \frac{3}{5}) + 2 πn

Show solutions in decimal form

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = 2.21429 \dots + 2 πn, x = - 2.21429 \dots + 2 πn

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