What is the general solution for -2sin(x)+cos(2x)=0 ?

The general solution for -2sin(x)+cos(2x)=0 is x=0.37473…+2pin,x=pi-0.37473…+2pin

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

Solution

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

Solution

x = 0.37473 \dots + 2 πn, x = π - 0.37473 \dots + 2 πn

Degrees

x = 21.47070 \dots^{\circ} + 36 0^{\circ} n, x = 158.52929 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

cos (2 x) - 2 sin (x)

Use the Double Angle identity:

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

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

1 - 2 sin (x) - 2 sin^{2} (x) = 0

Solve by substitution

1 - 2 sin (x) - 2 sin^{2} (x) = 0

Let:

sin (x) = u

1 - 2 u - 2 u^{2} = 0

1 - 2 u - 2 u^{2} = 0 : u = - \frac{1 + 3}{2}, u = \frac{3 - 1}{2}

1 - 2 u - 2 u^{2} = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 2, b = - 2, c = 1

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

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

(- 2)^{2} - 4 (- 2) \cdot 1 = 23

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

Apply rule

- (- a) = a

= (- 2)^{2} + 4 \cdot 2 \cdot 1

Apply exponent rule:

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

n

is even

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

= 2^{2} + 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 2^{2} + 8

2^{2} = 4

= 4 + 8

Add the numbers:

4 + 8 = 12

= 12

Prime factorization of

12 : 2^{2} \cdot 3

12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3

= 2^{2} \cdot 3

= 2^{2} \cdot 3

Apply radical rule:

= 3 2^{2}

Apply radical rule:

2^{2} = 2

= 23

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

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 2 3}{2 ( - 2 )}, u_{2} = \frac{- ( - 2 ) - 2 3}{2 ( - 2 )}

u = \frac{- ( - 2 ) + 2 3}{2 ( - 2 )} : - \frac{1 + 3}{2}

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

Remove parentheses:

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

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

Multiply the numbers:

2 \cdot 2 = 4

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

Apply the fraction rule:

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

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

Cancel

\frac{2 + 2 3}{4} : \frac{1 + 3}{2}

\frac{2 + 2 3}{4}

Factor

2 + 23 : 2 (1 + 3)

2 + 23

Rewrite as

= 2 \cdot 1 + 23

Factor out common term

2

= 2 (1 + 3)

= \frac{2 ( 1 + 3 )}{4}

Cancel the common factor:

2

= \frac{1 + 3}{2}

= - \frac{1 + 3}{2}

u = \frac{- ( - 2 ) - 2 3}{2 ( - 2 )} : \frac{3 - 1}{2}

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

Remove parentheses:

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

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

Multiply the numbers:

2 \cdot 2 = 4

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

Apply the fraction rule:

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

2 - 23 = - (23 - 2)

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

Factor

23 - 2 : 2 (3 - 1)

23 - 2

Rewrite as

= 23 - 2 \cdot 1

Factor out common term

2

= 2 (3 - 1)

= \frac{2 ( 3 - 1 )}{4}

Cancel the common factor:

2

= \frac{3 - 1}{2}

The solutions to the quadratic equation are:

u = - \frac{1 + 3}{2}, u = \frac{3 - 1}{2}

Substitute back

u = sin (x)

sin (x) = - \frac{1 + 3}{2}, sin (x) = \frac{3 - 1}{2}

sin (x) = - \frac{1 + 3}{2}, sin (x) = \frac{3 - 1}{2}

sin (x) = - \frac{1 + 3}{2} :

No Solution

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

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = \frac{3 - 1}{2} : x = arcsin (\frac{3 - 1}{2}) + 2 πn, x = π - arcsin (\frac{3 - 1}{2}) + 2 πn

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

Apply trig inverse properties

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

General solutions for

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

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

x = arcsin (\frac{3 - 1}{2}) + 2 πn, x = π - arcsin (\frac{3 - 1}{2}) + 2 πn

x = arcsin (\frac{3 - 1}{2}) + 2 πn, x = π - arcsin (\frac{3 - 1}{2}) + 2 πn

Combine all the solutions

x = arcsin (\frac{3 - 1}{2}) + 2 πn, x = π - arcsin (\frac{3 - 1}{2}) + 2 πn

Show solutions in decimal form

x = 0.37473 \dots + 2 πn, x = π - 0.37473 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for -2sin(x)+cos(2x)=0 ?
The general solution for -2sin(x)+cos(2x)=0 is x=0.37473…+2pin,x=pi-0.37473…+2pin