What is the general solution for 2cos(2x)+4sqrt(3)sin(x)=5 ?

The general solution for 2cos(2x)+4sqrt(3)sin(x)=5 is x= pi/3+2pin,x=(2pi)/3+2pin

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2cos(2x)+4sqrt(3)sin(x)=5

Solution

2 cos (2 x) + 43 sin (x) = 5

Solution

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

Degrees

x = 6 0^{\circ} + 36 0^{\circ} n, x = 12 0^{\circ} + 36 0^{\circ} n

Solution steps

2 cos (2 x) + 43 sin (x) = 5

Subtract

5

from both sides

2 cos (2 x) + 43 sin (x) - 5 = 0

Rewrite using trig identities

- 5 + 2 cos (2 x) + 4 sin (x) 3

Use the Double Angle identity:

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

= - 5 + 2 (1 - 2 sin^{2} (x)) + 43 sin (x)

Simplify

- 5 + 2 (1 - 2 sin^{2} (x)) + 43 sin (x) : 43 sin (x) - 4 sin^{2} (x) - 3

- 5 + 2 (1 - 2 sin^{2} (x)) + 43 sin (x)

Expand

2 (1 - 2 sin^{2} (x)) : 2 - 4 sin^{2} (x)

2 (1 - 2 sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

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

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

Simplify

2 \cdot 1 - 2 \cdot 2 sin^{2} (x) : 2 - 4 sin^{2} (x)

2 \cdot 1 - 2 \cdot 2 sin^{2} (x)

Multiply the numbers:

2 \cdot 1 = 2

= 2 - 2 \cdot 2 sin^{2} (x)

Multiply the numbers:

2 \cdot 2 = 4

= 2 - 4 sin^{2} (x)

= 2 - 4 sin^{2} (x)

= - 5 + 2 - 4 sin^{2} (x) + 43 sin (x)

Add/Subtract the numbers:

- 5 + 2 = - 3

= 43 sin (x) - 4 sin^{2} (x) - 3

= 43 sin (x) - 4 sin^{2} (x) - 3

- 3 - 4 sin^{2} (x) + 4 sin (x) 3 = 0

Solve by substitution

- 3 - 4 sin^{2} (x) + 4 sin (x) 3 = 0

Let:

sin (x) = u

- 3 - 4 u^{2} + 4 u 3 = 0

- 3 - 4 u^{2} + 4 u 3 = 0 : u = \frac{3}{2}

- 3 - 4 u^{2} + 4 u 3 = 0

Write in the standard form

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

- 4 u^{2} + 43 u - 3 = 0

Solve with the quadratic formula

- 4 u^{2} + 43 u - 3 = 0

Quadratic Equation Formula:

For

a = - 4, b = 43, c = - 3

u_{1, 2} = \frac{- 4 3 \pm ( 4 3 ) ^{2} - 4 ( - 4 ) ( - 3 )}{2 ( - 4 )}

u_{1, 2} = \frac{- 4 3 \pm ( 4 3 ) ^{2} - 4 ( - 4 ) ( - 3 )}{2 ( - 4 )}

(43)^{2} - 4 (- 4) (- 3) = 0

(43)^{2} - 4 (- 4) (- 3)

Apply rule

- (- a) = a

= (43)^{2} - 4 \cdot 4 \cdot 3

(43)^{2} = 4^{2} \cdot 3

(43)^{2}

Apply exponent rule:

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

= 4^{2} (3)^{2}

(3)^{2} : 3

Apply radical rule:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

Apply exponent rule:

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

= 3^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

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

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 3

= 4^{2} \cdot 3

4 \cdot 4 \cdot 3 = 48

4 \cdot 4 \cdot 3

Multiply the numbers:

4 \cdot 4 \cdot 3 = 48

= 48

= 4^{2} \cdot 3 - 48

4^{2} \cdot 3 = 48

4^{2} \cdot 3

4^{2} = 16

= 16 \cdot 3

Multiply the numbers:

16 \cdot 3 = 48

= 48

= 48 - 48

Subtract the numbers:

48 - 48 = 0

= 0

u_{1, 2} = \frac{- 4 3 \pm 0}{2 ( - 4 )}

u = \frac{- 4 3}{2 ( - 4 )}

\frac{- 4 3}{2 ( - 4 )} = \frac{3}{2}

\frac{- 4 3}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

= \frac{- 4 3}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 4 3}{- 8}

Apply the fraction rule:

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

= \frac{4 3}{8}

Cancel the common factor:

4

= \frac{3}{2}

u = \frac{3}{2}

The solution to the quadratic equation is:

u = \frac{3}{2}

Substitute back

u = sin (x)

sin (x) = \frac{3}{2}

sin (x) = \frac{3}{2}

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

sin (x) = \frac{3}{2}

General solutions for

sin (x) = \frac{3}{2}

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

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

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

Combine all the solutions

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

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos(2x)+4sqrt(3)sin(x)=5 ?
The general solution for 2cos(2x)+4sqrt(3)sin(x)=5 is x= pi/3+2pin,x=(2pi)/3+2pin