What is the general solution for 4sin(x)-cos(2x)=1 ?

The general solution for 4sin(x)-cos(2x)=1 is x=0.42707…+2pin,x=pi-0.42707…+2pin

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4sin(x)-cos(2x)=1

Solution

4 sin (x) - cos (2 x) = 1

Solution

x = 0.42707 \dots + 2 πn, x = π - 0.42707 \dots + 2 πn

Degrees

x = 24.46980 \dots^{\circ} + 36 0^{\circ} n, x = 155.53019 \dots^{\circ} + 36 0^{\circ} n

Solution steps

4 sin (x) - cos (2 x) = 1

Subtract

1

from both sides

4 sin (x) - cos (2 x) - 1 = 0

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Simplify

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

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

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

Subtract the numbers:

- 1 - 1 = - 2

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

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

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

Solve by substitution

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

Let:

sin (x) = u

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

- 2 + 2 u^{2} + 4 u = 0 : u = - 1 + 2, u = - 1 - 2

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 2, b = 4, c = - 2

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

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

4^{2} - 4 \cdot 2 (- 2) = 42

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 4^{2} + 16

4^{2} = 16

= 16 + 16

Add the numbers:

16 + 16 = 32

= 32

Prime factorization of

32 : 2^{5}

32

32

divides by

232 = 16 \cdot 2

= 2 \cdot 16

16

divides by

216 = 8 \cdot 2

= 2 \cdot 2 \cdot 8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

= 2^{5}

= 2^{5}

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{4} \cdot 2

Apply radical rule:

= 2 2^{4}

Apply radical rule:

2^{4} = 2^{\frac{4}{2}} = 2^{2}

= 2^{2} 2

Refine

= 42

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

Separate the solutions

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

u = \frac{- 4 + 4 2}{2 \cdot 2} : - 1 + 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4 + 4 2}{4}

Factor

- 4 + 42 : 4 (- 1 + 2)

- 4 + 42

Rewrite as

= - 4 \cdot 1 + 42

Factor out common term

4

= 4 (- 1 + 2)

= \frac{4 ( - 1 + 2 )}{4}

Divide the numbers:

\frac{4}{4} = 1

= - 1 + 2

u = \frac{- 4 - 4 2}{2 \cdot 2} : - 1 - 2

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

Multiply the numbers:

2 \cdot 2 = 4

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

Factor

- 4 - 42 : - 4 (1 + 2)

- 4 - 42

Rewrite as

= - 4 \cdot 1 - 42

Factor out common term

4

= - 4 (1 + 2)

= - \frac{4 ( 1 + 2 )}{4}

Divide the numbers:

\frac{4}{4} = 1

= - (1 + 2)

Negate

- (1 + 2) = - 1 - 2

= - 1 - 2

The solutions to the quadratic equation are:

u = - 1 + 2, u = - 1 - 2

Substitute back

u = sin (x)

sin (x) = - 1 + 2, sin (x) = - 1 - 2

sin (x) = - 1 + 2, sin (x) = - 1 - 2

sin (x) = - 1 + 2 : x = arcsin (- 1 + 2) + 2 πn, x = π - arcsin (- 1 + 2) + 2 πn

sin (x) = - 1 + 2

Apply trig inverse properties

sin (x) = - 1 + 2

General solutions for

sin (x) = - 1 + 2

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

x = arcsin (- 1 + 2) + 2 πn, x = π - arcsin (- 1 + 2) + 2 πn

x = arcsin (- 1 + 2) + 2 πn, x = π - arcsin (- 1 + 2) + 2 πn

sin (x) = - 1 - 2 :

No Solution

sin (x) = - 1 - 2

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = arcsin (- 1 + 2) + 2 πn, x = π - arcsin (- 1 + 2) + 2 πn

Show solutions in decimal form

x = 0.42707 \dots + 2 πn, x = π - 0.42707 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4sin(x)-cos(2x)=1 ?
The general solution for 4sin(x)-cos(2x)=1 is x=0.42707…+2pin,x=pi-0.42707…+2pin