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

The general solution for 5sin(x)-4=2cos(2x) is x=0.84806…+2pin,x=pi-0.84806…+2pin

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5sin(x)-4=2cos(2x)

Solution

5 sin (x) - 4 = 2 cos (2 x)

Solution

x = 0.84806 \dots + 2 πn, x = π - 0.84806 \dots + 2 πn

Degrees

x = 48.59037 \dots^{\circ} + 36 0^{\circ} n, x = 131.40962 \dots^{\circ} + 36 0^{\circ} n

Solution steps

5 sin (x) - 4 = 2 cos (2 x)

Subtract

2 cos (2 x)

from both sides

5 sin (x) - 4 - 2 cos (2 x) = 0

Rewrite using trig identities

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

Use the Double Angle identity:

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

= - 4 - 2 (1 - 2 sin^{2} (x)) + 5 sin (x)

Simplify

- 4 - 2 (1 - 2 sin^{2} (x)) + 5 sin (x) : 4 sin^{2} (x) + 5 sin (x) - 6

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

Apply minus-plus rules

- (- a) = a

= - 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)

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

Subtract the numbers:

- 4 - 2 = - 6

= 4 sin^{2} (x) + 5 sin (x) - 6

= 4 sin^{2} (x) + 5 sin (x) - 6

- 6 + 4 sin^{2} (x) + 5 sin (x) = 0

Solve by substitution

- 6 + 4 sin^{2} (x) + 5 sin (x) = 0

Let:

sin (x) = u

- 6 + 4 u^{2} + 5 u = 0

- 6 + 4 u^{2} + 5 u = 0 : u = \frac{3}{4}, u = - 2

- 6 + 4 u^{2} + 5 u = 0

Write in the standard form

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

4 u^{2} + 5 u - 6 = 0

Solve with the quadratic formula

4 u^{2} + 5 u - 6 = 0

Quadratic Equation Formula:

For

a = 4, b = 5, c = - 6

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 4 ( - 6 )}{2 \cdot 4}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 4 ( - 6 )}{2 \cdot 4}

5^{2} - 4 \cdot 4 (- 6) = 11

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

Apply rule

- (- a) = a

= 5^{2} + 4 \cdot 4 \cdot 6

Multiply the numbers:

4 \cdot 4 \cdot 6 = 96

= 5^{2} + 96

5^{2} = 25

= 25 + 96

Add the numbers:

25 + 96 = 121

= 121

Factor the number:

121 = 1 1^{2}

= 1 1^{2}

Apply radical rule:

1 1^{2} = 11

= 11

u_{1, 2} = \frac{- 5 \pm 11}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- 5 + 11}{2 \cdot 4}, u_{2} = \frac{- 5 - 11}{2 \cdot 4}

u = \frac{- 5 + 11}{2 \cdot 4} : \frac{3}{4}

\frac{- 5 + 11}{2 \cdot 4}

Add/Subtract the numbers:

- 5 + 11 = 6

= \frac{6}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{6}{8}

Cancel the common factor:

2

= \frac{3}{4}

u = \frac{- 5 - 11}{2 \cdot 4} : - 2

\frac{- 5 - 11}{2 \cdot 4}

Subtract the numbers:

- 5 - 11 = - 16

= \frac{- 16}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 16}{8}

Apply the fraction rule:

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

= - \frac{16}{8}

Divide the numbers:

\frac{16}{8} = 2

= - 2

The solutions to the quadratic equation are:

u = \frac{3}{4}, u = - 2

Substitute back

u = sin (x)

sin (x) = \frac{3}{4}, sin (x) = - 2

sin (x) = \frac{3}{4}, sin (x) = - 2

sin (x) = \frac{3}{4} : x = arcsin (\frac{3}{4}) + 2 πn, x = π - arcsin (\frac{3}{4}) + 2 πn

sin (x) = \frac{3}{4}

Apply trig inverse properties

sin (x) = \frac{3}{4}

General solutions for

sin (x) = \frac{3}{4}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{3}{4}) + 2 πn, x = π - arcsin (\frac{3}{4}) + 2 πn

x = arcsin (\frac{3}{4}) + 2 πn, x = π - arcsin (\frac{3}{4}) + 2 πn

sin (x) = - 2 :

No Solution

sin (x) = - 2

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = arcsin (\frac{3}{4}) + 2 πn, x = π - arcsin (\frac{3}{4}) + 2 πn

Show solutions in decimal form

x = 0.84806 \dots + 2 πn, x = π - 0.84806 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 5sin(x)-4=2cos(2x) ?
The general solution for 5sin(x)-4=2cos(2x) is x=0.84806…+2pin,x=pi-0.84806…+2pin