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4cos^2(x)+cos(2x)-7cos(x)=-2,0<= x<= 360

Solution

4 cos^{2} (x) + cos (2 x) - 7 cos (x) = - 2, 0^{\circ} \leq x \leq 36 0^{\circ}

Solution

x = 0, x = 36 0^{\circ}, x = 1.40334 \dots, x = 36 0^{\circ} - 1.40334 \dots

Radians

x = 0, x = 2 π, x = 1.40334 \dots, x = 2 π - 1.40334 \dots

Solution steps

4 cos^{2} (x) + cos (2 x) - 7 cos (x) = - 2, 0^{\circ} \leq x \leq 36 0^{\circ}

Subtract

- 2

from both sides

4 cos^{2} (x) + cos (2 x) - 7 cos (x) + 2 = 0

Rewrite using trig identities

2 + cos (2 x) + 4 cos^{2} (x) - 7 cos (x)

Use the Double Angle identity:

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

= 2 + 2 cos^{2} (x) - 1 + 4 cos^{2} (x) - 7 cos (x)

Simplify

2 + 2 cos^{2} (x) - 1 + 4 cos^{2} (x) - 7 cos (x) : 6 cos^{2} (x) - 7 cos (x) + 1

2 + 2 cos^{2} (x) - 1 + 4 cos^{2} (x) - 7 cos (x)

Group like terms

= 2 cos^{2} (x) + 4 cos^{2} (x) - 7 cos (x) + 2 - 1

Add similar elements:

2 cos^{2} (x) + 4 cos^{2} (x) = 6 cos^{2} (x)

= 6 cos^{2} (x) - 7 cos (x) + 2 - 1

Add/Subtract the numbers:

2 - 1 = 1

= 6 cos^{2} (x) - 7 cos (x) + 1

= 6 cos^{2} (x) - 7 cos (x) + 1

1 + 6 cos^{2} (x) - 7 cos (x) = 0

Solve by substitution

1 + 6 cos^{2} (x) - 7 cos (x) = 0

Let:

cos (x) = u

1 + 6 u^{2} - 7 u = 0

1 + 6 u^{2} - 7 u = 0 : u = 1, u = \frac{1}{6}

1 + 6 u^{2} - 7 u = 0

Write in the standard form

a x^{2} + b x + c = 0

6 u^{2} - 7 u + 1 = 0

Solve with the quadratic formula

6 u^{2} - 7 u + 1 = 0

Quadratic Equation Formula:

For

a = 6, b = - 7, c = 1

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 6 \cdot 1}{2 \cdot 6}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 6 \cdot 1}{2 \cdot 6}

(- 7)^{2} - 4 \cdot 6 \cdot 1 = 5

(- 7)^{2} - 4 \cdot 6 \cdot 1

Apply exponent rule:

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

n

is even

(- 7)^{2} = 7^{2}

= 7^{2} - 4 \cdot 6 \cdot 1

Multiply the numbers:

4 \cdot 6 \cdot 1 = 24

= 7^{2} - 24

7^{2} = 49

= 49 - 24

Subtract the numbers:

49 - 24 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- ( - 7 ) \pm 5}{2 \cdot 6}

Separate the solutions

u_{1} = \frac{- ( - 7 ) + 5}{2 \cdot 6}, u_{2} = \frac{- ( - 7 ) - 5}{2 \cdot 6}

u = \frac{- ( - 7 ) + 5}{2 \cdot 6} : 1

\frac{- ( - 7 ) + 5}{2 \cdot 6}

Apply rule

- (- a) = a

= \frac{7 + 5}{2 \cdot 6}

Add the numbers:

7 + 5 = 12

= \frac{12}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{12}{12}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 7 ) - 5}{2 \cdot 6} : \frac{1}{6}

\frac{- ( - 7 ) - 5}{2 \cdot 6}

Apply rule

- (- a) = a

= \frac{7 - 5}{2 \cdot 6}

Subtract the numbers:

7 - 5 = 2

= \frac{2}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{2}{12}

Cancel the common factor:

2

= \frac{1}{6}

The solutions to the quadratic equation are:

u = 1, u = \frac{1}{6}

Substitute back

u = cos (x)

cos (x) = 1, cos (x) = \frac{1}{6}

cos (x) = 1, cos (x) = \frac{1}{6}

cos (x) = 1, 0 \leq x \leq 36 0^{\circ} : x = 0, x = 36 0^{\circ}

cos (x) = 1, 0 \leq x \leq 36 0^{\circ}

General solutions for

cos (x) = 1

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

Solve

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

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

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

x = 36 0^{\circ} n

x = 36 0^{\circ} n

Solutions for the range

0 \leq x \leq 36 0^{\circ}

x = 0, x = 36 0^{\circ}

cos (x) = \frac{1}{6}, 0 \leq x \leq 36 0^{\circ} : x = arccos (\frac{1}{6}), x = 36 0^{\circ} - arccos (\frac{1}{6})

cos (x) = \frac{1}{6}, 0 \leq x \leq 36 0^{\circ}

Apply trig inverse properties

cos (x) = \frac{1}{6}

General solutions for

cos (x) = \frac{1}{6}

cos (x) = a \Rightarrow x = arccos (a) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (a) + 36 0^{\circ} n

x = arccos (\frac{1}{6}) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (\frac{1}{6}) + 36 0^{\circ} n

x = arccos (\frac{1}{6}) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (\frac{1}{6}) + 36 0^{\circ} n

Solutions for the range

0 \leq x \leq 36 0^{\circ}

x = arccos (\frac{1}{6}), x = 36 0^{\circ} - arccos (\frac{1}{6})

Combine all the solutions

x = 0, x = 36 0^{\circ}, x = arccos (\frac{1}{6}), x = 36 0^{\circ} - arccos (\frac{1}{6})

Show solutions in decimal form

x = 0, x = 36 0^{\circ}, x = 1.40334 \dots, x = 36 0^{\circ} - 1.40334 \dots

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