What is the general solution for 10+5cos(2x)=15cos(x) ?

The general solution for 10+5cos(2x)=15cos(x) is x=2pin,x= pi/3+2pin,x=(5pi)/3+2pin

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Popular Trigonometry >

10+5cos(2x)=15cos(x)

Solution

10 + 5 cos (2 x) = 15 cos (x)

Solution

x = 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

10 + 5 cos (2 x) = 15 cos (x)

Subtract

15 cos (x)

from both sides

10 + 5 cos (2 x) - 15 cos (x) = 0

Rewrite using trig identities

10 - 15 cos (x) + 5 cos (2 x)

Use the Double Angle identity:

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

= 10 - 15 cos (x) + 5 (2 cos^{2} (x) - 1)

Simplify

10 - 15 cos (x) + 5 (2 cos^{2} (x) - 1) : 10 cos^{2} (x) - 15 cos (x) + 5

10 - 15 cos (x) + 5 (2 cos^{2} (x) - 1)

Expand

5 (2 cos^{2} (x) - 1) : 10 cos^{2} (x) - 5

5 (2 cos^{2} (x) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 5, b = 2 cos^{2} (x), c = 1

= 5 \cdot 2 cos^{2} (x) - 5 \cdot 1

Simplify

5 \cdot 2 cos^{2} (x) - 5 \cdot 1 : 10 cos^{2} (x) - 5

5 \cdot 2 cos^{2} (x) - 5 \cdot 1

Multiply the numbers:

5 \cdot 2 = 10

= 10 cos^{2} (x) - 5 \cdot 1

Multiply the numbers:

5 \cdot 1 = 5

= 10 cos^{2} (x) - 5

= 10 cos^{2} (x) - 5

= 10 - 15 cos (x) + 10 cos^{2} (x) - 5

Simplify

10 - 15 cos (x) + 10 cos^{2} (x) - 5 : 10 cos^{2} (x) - 15 cos (x) + 5

10 - 15 cos (x) + 10 cos^{2} (x) - 5

Group like terms

= - 15 cos (x) + 10 cos^{2} (x) + 10 - 5

Add/Subtract the numbers:

10 - 5 = 5

= 10 cos^{2} (x) - 15 cos (x) + 5

= 10 cos^{2} (x) - 15 cos (x) + 5

= 10 cos^{2} (x) - 15 cos (x) + 5

5 + 10 cos^{2} (x) - 15 cos (x) = 0

Solve by substitution

5 + 10 cos^{2} (x) - 15 cos (x) = 0

Let:

cos (x) = u

5 + 10 u^{2} - 15 u = 0

5 + 10 u^{2} - 15 u = 0 : u = 1, u = \frac{1}{2}

5 + 10 u^{2} - 15 u = 0

Write in the standard form

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

10 u^{2} - 15 u + 5 = 0

Solve with the quadratic formula

10 u^{2} - 15 u + 5 = 0

Quadratic Equation Formula:

For

a = 10, b = - 15, c = 5

u_{1, 2} = \frac{- ( - 15 ) \pm ( - 15 ) ^{2} - 4 \cdot 10 \cdot 5}{2 \cdot 10}

u_{1, 2} = \frac{- ( - 15 ) \pm ( - 15 ) ^{2} - 4 \cdot 10 \cdot 5}{2 \cdot 10}

(- 15)^{2} - 4 \cdot 10 \cdot 5 = 5

(- 15)^{2} - 4 \cdot 10 \cdot 5

Apply exponent rule:

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

n

is even

(- 15)^{2} = 1 5^{2}

= 1 5^{2} - 4 \cdot 10 \cdot 5

Multiply the numbers:

4 \cdot 10 \cdot 5 = 200

= 1 5^{2} - 200

1 5^{2} = 225

= 225 - 200

Subtract the numbers:

225 - 200 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

n a^{n} = a

5^{2} = 5

= 5

u_{1, 2} = \frac{- ( - 15 ) \pm 5}{2 \cdot 10}

Separate the solutions

u_{1} = \frac{- ( - 15 ) + 5}{2 \cdot 10}, u_{2} = \frac{- ( - 15 ) - 5}{2 \cdot 10}

u = \frac{- ( - 15 ) + 5}{2 \cdot 10} : 1

\frac{- ( - 15 ) + 5}{2 \cdot 10}

Apply rule

- (- a) = a

= \frac{15 + 5}{2 \cdot 10}

Add the numbers:

15 + 5 = 20

= \frac{20}{2 \cdot 10}

Multiply the numbers:

2 \cdot 10 = 20

= \frac{20}{20}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 15 ) - 5}{2 \cdot 10} : \frac{1}{2}

\frac{- ( - 15 ) - 5}{2 \cdot 10}

Apply rule

- (- a) = a

= \frac{15 - 5}{2 \cdot 10}

Subtract the numbers:

15 - 5 = 10

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

Multiply the numbers:

2 \cdot 10 = 20

= \frac{10}{20}

Cancel the common factor:

10

= \frac{1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

cos (x) = 1, cos (x) = \frac{1}{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} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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

General solutions for

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

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}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Combine all the solutions

x = 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 10+5cos(2x)=15cos(x) ?
The general solution for 10+5cos(2x)=15cos(x) is x=2pin,x= pi/3+2pin,x=(5pi)/3+2pin