What is the general solution for cos(4x)+1=3sin(2x),-pi/2 <= x<= pi/2 ?

Q: What is the general solution for cos(4x)+1=3sin(2x),-pi/2 <= x<= pi/2 ?

The general solution for cos(4x)+1=3sin(2x),-pi/2 <= x<= pi/2 is x= pi/(12),x=(5pi)/(12)

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cos(4x)+1=3sin(2x),-pi/2 <= x<= pi/2

Solution

cos (4 x) + 1 = 3 sin (2 x), - \frac{π}{2} \leq x \leq \frac{π}{2}

Solution

x = \frac{π}{12}, x = \frac{5 π}{12}

Degrees

x = 1 5^{\circ}, x = 7 5^{\circ}

Solution steps

cos (4 x) + 1 = 3 sin (2 x), - \frac{π}{2} \leq x \leq \frac{π}{2}

Subtract

3 sin (2 x)

from both sides

cos (4 x) + 1 - 3 sin (2 x) = 0

Rewrite using trig identities

1 + cos (4 x) - 3 sin (2 x)

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

cos (4 x)

Rewrite as

= cos (2 \cdot 2 x)

Use the Double Angle identity:

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

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

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

= 1 + 2 cos^{2} (2 x) - 1 - 3 sin (2 x)

Simplify

1 + 2 cos^{2} (2 x) - 1 - 3 sin (2 x) : 2 cos^{2} (2 x) - 3 sin (2 x)

1 + 2 cos^{2} (2 x) - 1 - 3 sin (2 x)

Group like terms

= 2 cos^{2} (2 x) - 3 sin (2 x) + 1 - 1

1 - 1 = 0

= 2 cos^{2} (2 x) - 3 sin (2 x)

= 2 cos^{2} (2 x) - 3 sin (2 x)

Use the Pythagorean identity:

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

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

= 2 (1 - sin^{2} (2 x)) - 3 sin (2 x)

(1 - sin^{2} (2 x)) \cdot 2 - 3 sin (2 x) = 0

Solve by substitution

(1 - sin^{2} (2 x)) \cdot 2 - 3 sin (2 x) = 0

Let:

sin (2 x) = u

(1 - u^{2}) \cdot 2 - 3 u = 0

(1 - u^{2}) \cdot 2 - 3 u = 0 : u = - 2, u = \frac{1}{2}

(1 - u^{2}) \cdot 2 - 3 u = 0

Expand

(1 - u^{2}) \cdot 2 - 3 u : 2 - 2 u^{2} - 3 u

(1 - u^{2}) \cdot 2 - 3 u

= 2 (1 - u^{2}) - 3 u

Expand

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

2 (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

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

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

Multiply the numbers:

2 \cdot 1 = 2

= 2 - 2 u^{2}

= 2 - 2 u^{2} - 3 u

2 - 2 u^{2} - 3 u = 0

Write in the standard form

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

- 2 u^{2} - 3 u + 2 = 0

Solve with the quadratic formula

- 2 u^{2} - 3 u + 2 = 0

Quadratic Equation Formula:

For

a = - 2, b = - 3, c = 2

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 2 ) \cdot 2}{2 ( - 2 )}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 2 ) \cdot 2}{2 ( - 2 )}

(- 3)^{2} - 4 (- 2) \cdot 2 = 5

(- 3)^{2} - 4 (- 2) \cdot 2

Apply rule

- (- a) = a

= (- 3)^{2} + 4 \cdot 2 \cdot 2

Apply exponent rule:

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

n

is even

(- 3)^{2} = 3^{2}

= 3^{2} + 4 \cdot 2 \cdot 2

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 3^{2} + 16

3^{2} = 9

= 9 + 16

Add the numbers:

9 + 16 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- ( - 3 ) \pm 5}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- ( - 3 ) + 5}{2 ( - 2 )}, u_{2} = \frac{- ( - 3 ) - 5}{2 ( - 2 )}

u = \frac{- ( - 3 ) + 5}{2 ( - 2 )} : - 2

\frac{- ( - 3 ) + 5}{2 ( - 2 )}

Remove parentheses:

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

= \frac{3 + 5}{- 2 \cdot 2}

Add the numbers:

3 + 5 = 8

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{8}{- 4}

Apply the fraction rule:

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

= - \frac{8}{4}

Divide the numbers:

\frac{8}{4} = 2

= - 2

u = \frac{- ( - 3 ) - 5}{2 ( - 2 )} : \frac{1}{2}

\frac{- ( - 3 ) - 5}{2 ( - 2 )}

Remove parentheses:

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

= \frac{3 - 5}{- 2 \cdot 2}

Subtract the numbers:

3 - 5 = - 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

Apply the fraction rule:

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

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = sin (2 x)

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

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

sin (2 x) = - 2, - \frac{π}{2} \leq x \leq \frac{π}{2} :

No Solution

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

- 1 \leq sin (x) \leq 1

No Solution

sin (2 x) = \frac{1}{2}, - \frac{π}{2} \leq x \leq \frac{π}{2} : x = \frac{π}{12}, x = \frac{5 π}{12}

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

General solutions for

sin (2 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}

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

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

Solve

2 x = \frac{π}{6} + 2 πn : x = \frac{π}{12} + πn

2 x = \frac{π}{6} + 2 πn

Divide both sides by

2

2 x = \frac{π}{6} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{π}{6}}{2} + \frac{2 πn}{2} : \frac{π}{12} + πn

\frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{6}}{2} = \frac{π}{12}

\frac{\frac{π}{6}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{12} + πn

x = \frac{π}{12} + πn

x = \frac{π}{12} + πn

x = \frac{π}{12} + πn

Solve

2 x = \frac{5 π}{6} + 2 πn : x = \frac{5 π}{12} + πn

2 x = \frac{5 π}{6} + 2 πn

Divide both sides by

2

2 x = \frac{5 π}{6} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2} : \frac{5 π}{12} + πn

\frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{6}}{2} = \frac{5 π}{12}

\frac{\frac{5 π}{6}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{5 π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{5 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

Solutions for the range

- \frac{π}{2} \leq x \leq \frac{π}{2}

x = \frac{π}{12}, x = \frac{5 π}{12}

Combine all the solutions

x = \frac{π}{12}, x = \frac{5 π}{12}

Popular Examples

3-6cos(x+pi/3)=0

3 - 6 cos (x + \frac{π}{3}) = 0

sin(7.07t)=0.5

sin (7.07 t) = 0.5

tan(x)= 1/(sqrt(3)),0<= x<2pi

tan (x) = \frac{1}{3}, 0 \leq x < 2 π

-2sin(2x)=2pi

- 2 sin (2 x) = 2 π

arccos(θ)= 17/18

arccos (θ) = \frac{17}{18}

Frequently Asked Questions (FAQ)

What is the general solution for cos(4x)+1=3sin(2x),-pi/2 <= x<= pi/2 ?
The general solution for cos(4x)+1=3sin(2x),-pi/2 <= x<= pi/2 is x= pi/(12),x=(5pi)/(12)