What is the general solution for -14sin(x)+14cos(2x)=0 ?

The general solution for -14sin(x)+14cos(2x)=0 is x=(3pi)/2+2pin,x= pi/6+2pin,x=(5pi)/6+2pin

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-14sin(x)+14cos(2x)=0

Solution

- 14 sin (x) + 14 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

- 14 sin (x) + 14 cos (2 x) = 0

Rewrite using trig identities

14 cos (2 x) - 14 sin (x)

Use the Double Angle identity:

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

= 14 (1 - 2 sin^{2} (x)) - 14 sin (x)

(1 - 2 sin^{2} (x)) \cdot 14 - 14 sin (x) = 0

Solve by substitution

(1 - 2 sin^{2} (x)) \cdot 14 - 14 sin (x) = 0

Let:

sin (x) = u

(1 - 2 u^{2}) \cdot 14 - 14 u = 0

(1 - 2 u^{2}) \cdot 14 - 14 u = 0 : u = - 1, u = \frac{1}{2}

(1 - 2 u^{2}) \cdot 14 - 14 u = 0

Expand

(1 - 2 u^{2}) \cdot 14 - 14 u : 14 - 28 u^{2} - 14 u

(1 - 2 u^{2}) \cdot 14 - 14 u

= 14 (1 - 2 u^{2}) - 14 u

Expand

14 (1 - 2 u^{2}) : 14 - 28 u^{2}

14 (1 - 2 u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = 14, b = 1, c = 2 u^{2}

= 14 \cdot 1 - 14 \cdot 2 u^{2}

Simplify

14 \cdot 1 - 14 \cdot 2 u^{2} : 14 - 28 u^{2}

14 \cdot 1 - 14 \cdot 2 u^{2}

Multiply the numbers:

14 \cdot 1 = 14

= 14 - 14 \cdot 2 u^{2}

Multiply the numbers:

14 \cdot 2 = 28

= 14 - 28 u^{2}

= 14 - 28 u^{2}

= 14 - 28 u^{2} - 14 u

14 - 28 u^{2} - 14 u = 0

Write in the standard form

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

- 28 u^{2} - 14 u + 14 = 0

Solve with the quadratic formula

- 28 u^{2} - 14 u + 14 = 0

Quadratic Equation Formula:

For

a = - 28, b = - 14, c = 14

u_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 ( - 28 ) \cdot 14}{2 ( - 28 )}

u_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 ( - 28 ) \cdot 14}{2 ( - 28 )}

(- 14)^{2} - 4 (- 28) \cdot 14 = 42

(- 14)^{2} - 4 (- 28) \cdot 14

Apply rule

- (- a) = a

= (- 14)^{2} + 4 \cdot 28 \cdot 14

Apply exponent rule:

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

n

is even

(- 14)^{2} = 1 4^{2}

= 1 4^{2} + 4 \cdot 28 \cdot 14

Multiply the numbers:

4 \cdot 28 \cdot 14 = 1568

= 1 4^{2} + 1568

1 4^{2} = 196

= 196 + 1568

Add the numbers:

196 + 1568 = 1764

= 1764

Factor the number:

1764 = 4 2^{2}

= 4 2^{2}

Apply radical rule:

4 2^{2} = 42

= 42

u_{1, 2} = \frac{- ( - 14 ) \pm 42}{2 ( - 28 )}

Separate the solutions

u_{1} = \frac{- ( - 14 ) + 42}{2 ( - 28 )}, u_{2} = \frac{- ( - 14 ) - 42}{2 ( - 28 )}

u = \frac{- ( - 14 ) + 42}{2 ( - 28 )} : - 1

\frac{- ( - 14 ) + 42}{2 ( - 28 )}

Remove parentheses:

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

= \frac{14 + 42}{- 2 \cdot 28}

Add the numbers:

14 + 42 = 56

= \frac{56}{- 2 \cdot 28}

Multiply the numbers:

2 \cdot 28 = 56

= \frac{56}{- 56}

Apply the fraction rule:

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

= - \frac{56}{56}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 14 ) - 42}{2 ( - 28 )} : \frac{1}{2}

\frac{- ( - 14 ) - 42}{2 ( - 28 )}

Remove parentheses:

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

= \frac{14 - 42}{- 2 \cdot 28}

Subtract the numbers:

14 - 42 = - 28

= \frac{- 28}{- 2 \cdot 28}

Multiply the numbers:

2 \cdot 28 = 56

= \frac{- 28}{- 56}

Apply the fraction rule:

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

= \frac{28}{56}

Cancel the common factor:

28

= \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 -14sin(x)+14cos(2x)=0 ?
The general solution for -14sin(x)+14cos(2x)=0 is x=(3pi)/2+2pin,x= pi/6+2pin,x=(5pi)/6+2pin