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Popular >

solvefor x,0=-2sin(x)-4cos(2x)

Solution

solve for

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

Solution

x = 1.00296 \dots + 2 πn, x = π - 1.00296 \dots + 2 πn, x = - 0.63486 \dots + 2 πn, x = π + 0.63486 \dots + 2 πn

Degrees

x = 57.46577 \dots^{\circ} + 36 0^{\circ} n, x = 122.53422 \dots^{\circ} + 36 0^{\circ} n, x = - 36.37519 \dots^{\circ} + 36 0^{\circ} n, x = 216.37519 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Switch sides

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

- (1 - 2 sin^{2} (x)) \cdot 4 - 2 sin (x) = 0

Solve by substitution

- (1 - 2 sin^{2} (x)) \cdot 4 - 2 sin (x) = 0

Let:

sin (x) = u

- (1 - 2 u^{2}) \cdot 4 - 2 u = 0

- (1 - 2 u^{2}) \cdot 4 - 2 u = 0 : u = \frac{1 + 33}{8}, u = \frac{1 - 33}{8}

- (1 - 2 u^{2}) \cdot 4 - 2 u = 0

Expand

- (1 - 2 u^{2}) \cdot 4 - 2 u : - 4 + 8 u^{2} - 2 u

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

= - 4 (1 - 2 u^{2}) - 2 u

Expand

- 4 (1 - 2 u^{2}) : - 4 + 8 u^{2}

- 4 (1 - 2 u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = - 4, b = 1, c = 2 u^{2}

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

Apply minus-plus rules

- (- a) = a

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

Simplify

- 4 \cdot 1 + 4 \cdot 2 u^{2} : - 4 + 8 u^{2}

- 4 \cdot 1 + 4 \cdot 2 u^{2}

Multiply the numbers:

4 \cdot 1 = 4

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

Multiply the numbers:

4 \cdot 2 = 8

= - 4 + 8 u^{2}

= - 4 + 8 u^{2}

= - 4 + 8 u^{2} - 2 u

- 4 + 8 u^{2} - 2 u = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 8, b = - 2, c = - 4

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 8 ( - 4 )}{2 \cdot 8}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 8 ( - 4 )}{2 \cdot 8}

(- 2)^{2} - 4 \cdot 8 (- 4) = 233

(- 2)^{2} - 4 \cdot 8 (- 4)

Apply rule

- (- a) = a

= (- 2)^{2} + 4 \cdot 8 \cdot 4

Apply exponent rule:

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

n

is even

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

= 2^{2} + 4 \cdot 8 \cdot 4

Multiply the numbers:

4 \cdot 8 \cdot 4 = 128

= 2^{2} + 128

2^{2} = 4

= 4 + 128

Add the numbers:

4 + 128 = 132

= 132

Prime factorization of

132 : 2^{2} \cdot 3 \cdot 11

132

132

divides by

2132 = 66 \cdot 2

= 2 \cdot 66

66

divides by

266 = 33 \cdot 2

= 2 \cdot 2 \cdot 33

33

divides by

333 = 11 \cdot 3

= 2 \cdot 2 \cdot 3 \cdot 11

2, 3, 11

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3 \cdot 11

= 2^{2} \cdot 3 \cdot 11

= 2^{2} \cdot 3 \cdot 11

Apply radical rule:

= 2^{2} 3 \cdot 11

Apply radical rule:

2^{2} = 2

= 2 3 \cdot 11

Refine

= 233

u_{1, 2} = \frac{- ( - 2 ) \pm 2 33}{2 \cdot 8}

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 2 33}{2 \cdot 8}, u_{2} = \frac{- ( - 2 ) - 2 33}{2 \cdot 8}

u = \frac{- ( - 2 ) + 2 33}{2 \cdot 8} : \frac{1 + 33}{8}

\frac{- ( - 2 ) + 2 33}{2 \cdot 8}

Apply rule

- (- a) = a

= \frac{2 + 2 33}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{2 + 2 33}{16}

Factor

2 + 233 : 2 (1 + 33)

2 + 233

Rewrite as

= 2 \cdot 1 + 233

Factor out common term

2

= 2 (1 + 33)

= \frac{2 ( 1 + 33 )}{16}

Cancel the common factor:

2

= \frac{1 + 33}{8}

u = \frac{- ( - 2 ) - 2 33}{2 \cdot 8} : \frac{1 - 33}{8}

\frac{- ( - 2 ) - 2 33}{2 \cdot 8}

Apply rule

- (- a) = a

= \frac{2 - 2 33}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{2 - 2 33}{16}

Factor

2 - 233 : 2 (1 - 33)

2 - 233

Rewrite as

= 2 \cdot 1 - 233

Factor out common term

2

= 2 (1 - 33)

= \frac{2 ( 1 - 33 )}{16}

Cancel the common factor:

2

= \frac{1 - 33}{8}

The solutions to the quadratic equation are:

u = \frac{1 + 33}{8}, u = \frac{1 - 33}{8}

Substitute back

u = sin (x)

sin (x) = \frac{1 + 33}{8}, sin (x) = \frac{1 - 33}{8}

sin (x) = \frac{1 + 33}{8}, sin (x) = \frac{1 - 33}{8}

sin (x) = \frac{1 + 33}{8} : x = arcsin (\frac{1 + 33}{8}) + 2 πn, x = π - arcsin (\frac{1 + 33}{8}) + 2 πn

sin (x) = \frac{1 + 33}{8}

Apply trig inverse properties

sin (x) = \frac{1 + 33}{8}

General solutions for

sin (x) = \frac{1 + 33}{8}

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

x = arcsin (\frac{1 + 33}{8}) + 2 πn, x = π - arcsin (\frac{1 + 33}{8}) + 2 πn

x = arcsin (\frac{1 + 33}{8}) + 2 πn, x = π - arcsin (\frac{1 + 33}{8}) + 2 πn

sin (x) = \frac{1 - 33}{8} : x = arcsin (\frac{1 - 33}{8}) + 2 πn, x = π + arcsin (- \frac{1 - 33}{8}) + 2 πn

sin (x) = \frac{1 - 33}{8}

Apply trig inverse properties

sin (x) = \frac{1 - 33}{8}

General solutions for

sin (x) = \frac{1 - 33}{8}

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

x = arcsin (\frac{1 - 33}{8}) + 2 πn, x = π + arcsin (- \frac{1 - 33}{8}) + 2 πn

x = arcsin (\frac{1 - 33}{8}) + 2 πn, x = π + arcsin (- \frac{1 - 33}{8}) + 2 πn

Combine all the solutions

x = arcsin (\frac{1 + 33}{8}) + 2 πn, x = π - arcsin (\frac{1 + 33}{8}) + 2 πn, x = arcsin (\frac{1 - 33}{8}) + 2 πn, x = π + arcsin (- \frac{1 - 33}{8}) + 2 πn

Show solutions in decimal form

x = 1.00296 \dots + 2 πn, x = π - 1.00296 \dots + 2 πn, x = - 0.63486 \dots + 2 πn, x = π + 0.63486 \dots + 2 πn

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