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

The general solution for 2(-sin(x)+cos(2x))=0 is x=(3pi)/2+2pin,x= pi/6+2pin,x=(5pi)/6+2pin

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

Solution

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

Solution

x = \frac{3 π}{2} + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Degrees

x = 27 0^{\circ} + 36 0^{\circ} n, x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

(cos (2 x) - sin (x)) \cdot 2

Use the Double Angle identity:

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

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

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

Solve by substitution

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

Let:

sin (x) = u

(1 - u - 2 u^{2}) \cdot 2 = 0

(1 - u - 2 u^{2}) \cdot 2 = 0 : u = - 1, u = \frac{1}{2}

(1 - u - 2 u^{2}) \cdot 2 = 0

Factor

(1 - u - 2 u^{2}) \cdot 2 : - 2 (u + 1) (2 u - 1)

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

Factor

- 2 u^{2} - u + 1 : - (2 u - 1) (u + 1)

- 2 u^{2} - u + 1

Factor out common term

- 1

= - (2 u^{2} + u - 1)

Factor

2 u^{2} + u - 1 : (2 u - 1) (u + 1)

2 u^{2} + u - 1

Write in the standard form

a x^{2} + b x + c

= 2 u^{2} + u - 1

Break the expression into groups

2 u^{2} + u - 1

Definition

Factors of

2 : 1, 2

2

Divisors (Factors)

Find the Prime factors of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Add 1

1

The factors of

2

1, 2

Negative factors of

2 : - 1, - 2

Multiply the factors by

- 1

to get the negative factors

- 1, - 2

For every two factors such that

u * v = - 2,

check if

u + v = 1

Check

u = 1, v = - 2 : u * v = - 2, u + v = - 1 \Rightarrow

FalseCheck

u = 2, v = - 1 : u * v = - 2, u + v = 1 \Rightarrow

True

u = 2, v = - 1

Group into

(a x^{2} + ux) + (vx + c)

(2 u^{2} - u) + (2 u - 1)

= (2 u^{2} - u) + (2 u - 1)

Factor out

u

from

2 u^{2} - u : u (2 u - 1)

2 u^{2} - u

Apply exponent rule:

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

u^{2} = uu

= 2 uu - u

Factor out common term

u

= u (2 u - 1)

= u (2 u - 1) + (2 u - 1)

Factor out common term

2 u - 1

= (2 u - 1) (u + 1)

= - (2 u - 1) (u + 1)

= - 2 (u + 1) (2 u - 1)

- 2 (u + 1) (2 u - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u + 1 = 0 or 2 u - 1 = 0

Solve

u + 1 = 0 : u = - 1

u + 1 = 0

Move

1

to the right side

u + 1 = 0

Subtract

1

from both sides

u + 1 - 1 = 0 - 1

Simplify

u = - 1

u = - 1

Solve

2 u - 1 = 0 : u = \frac{1}{2}

2 u - 1 = 0

Move

1

to the right side

2 u - 1 = 0

Add

1

to both sides

2 u - 1 + 1 = 0 + 1

Simplify

2 u = 1

2 u = 1

Divide both sides by

2

2 u = 1

Divide both sides by

2

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

Simplify

u = \frac{1}{2}

u = \frac{1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = sin (x)

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

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

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

sin (x) = - 1

General solutions for

sin (x) = - 1

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{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

General solutions for

sin (x) = \frac{1}{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}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Combine all the solutions

x = \frac{3 π}{2} + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2(-sin(x)+cos(2x))=0 ?
The general solution for 2(-sin(x)+cos(2x))=0 is x=(3pi)/2+2pin,x= pi/6+2pin,x=(5pi)/6+2pin