What is the general solution for cos(2x)sin(x)=1 ?

The general solution for cos(2x)sin(x)=1 is x=(3pi)/2+2pin

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cos(2x)sin(x)=1

Solution

cos (2 x) sin (x) = 1

Solution

x = \frac{3 π}{2} + 2 πn

Degrees

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

Solution steps

cos (2 x) sin (x) = 1

Subtract

1

from both sides

cos (2 x) sin (x) - 1 = 0

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Expand

- 1 + (1 - 2 u^{2}) u : - 1 + u - 2 u^{3}

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

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

Expand

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

u (1 - 2 u^{2})

Apply the distributive law:

a (b - c) = ab - a c

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

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

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

Simplify

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

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

1 \cdot u = u

1 \cdot u

Multiply:

1 \cdot u = u

= u

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

2 u^{2} u

Apply exponent rule:

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

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

= 2 u^{2 + 1}

Add the numbers:

2 + 1 = 3

= 2 u^{3}

= u - 2 u^{3}

= u - 2 u^{3}

= - 1 + u - 2 u^{3}

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

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

- 2 u^{3} + u - 1 = 0

Factor

- 2 u^{3} + u - 1 : - (u + 1) (2 u^{2} - 2 u + 1)

- 2 u^{3} + u - 1

Factor out common term

- 1

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

Factor

2 u^{3} - u + 1 : (u + 1) (2 u^{2} - 2 u + 1)

2 u^{3} - u + 1

Use the rational root theorem

a_{0} = 1, a_{n} = 2

The dividers of

a_{0} : 1,

The dividers of

a_{n} : 1, 2

Therefore, check the following rational numbers:

\pm \frac{1}{1 , 2}

- \frac{1}{1}

is a root of the expression, so factor out

u + 1

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

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

\frac{2 u ^{3} - u + 1}{u + 1}

Divide

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

Divide the leading coefficients of the numerator

2 u^{3} - u + 1

and the divisor

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

Quotient = 2 u^{2}

Multiply

u + 1

2 u^{2} : 2 u^{3} + 2 u^{2}

Subtract

2 u^{3} + 2 u^{2}

from

2 u^{3} - u + 1

to get new remainder

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

Therefore

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

= 2 u^{2} + \frac{- 2 u ^{2} - u + 1}{u + 1}

Divide

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

Divide the leading coefficients of the numerator

- 2 u^{2} - u + 1

and the divisor

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

Quotient = - 2 u

Multiply

u + 1

- 2 u : - 2 u^{2} - 2 u

Subtract

- 2 u^{2} - 2 u

from

- 2 u^{2} - u + 1

to get new remainder

Remainder = u + 1

Therefore

\frac{- 2 u ^{2} - u + 1}{u + 1} = - 2 u + \frac{u + 1}{u + 1}

= 2 u^{2} - 2 u + \frac{u + 1}{u + 1}

Divide

\frac{u + 1}{u + 1} : \frac{u + 1}{u + 1} = 1

Divide the leading coefficients of the numerator

u + 1

and the divisor

u + 1 : \frac{u}{u} = 1

Quotient = 1

Multiply

u + 1

1 : u + 1

Subtract

u + 1

from

u + 1

to get new remainder

Remainder = 0

Therefore

\frac{u + 1}{u + 1} = 1

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

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

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

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

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 \cdot 2 \cdot 1}{2 \cdot 2}

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

Simplify

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

(- 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

Apply imaginary number rule:

- a = i a

= i 8 - 2^{2}

- 2^{2} + 8 = 2

- 2^{2} + 8

2^{2} = 4

= - 4 + 8

Add/Subtract the numbers:

- 4 + 8 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= 2 i

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

Separate the solutions

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

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

\frac{- ( - 2 ) + 2 i}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{2 + 2 i}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2 + 2 i}{4}

Factor

2 + 2 i : 2 (1 + i)

2 + 2 i

Rewrite as

= 2 \cdot 1 + 2 i

Factor out common term

2

= 2 (1 + i)

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

Cancel the common factor:

2

= \frac{1 + i}{2}

Rewrite

\frac{1 + i}{2}

in standard complex form:

\frac{1}{2} + \frac{1}{2} i

\frac{1 + i}{2}

Apply the fraction rule:

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

\frac{1 + i}{2} = \frac{1}{2} + \frac{i}{2}

= \frac{1}{2} + \frac{i}{2}

= \frac{1}{2} + \frac{1}{2} i

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

\frac{- ( - 2 ) - 2 i}{2 \cdot 2}

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 2 = 4

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

Factor

2 - 2 i : 2 (1 - i)

2 - 2 i

Rewrite as

= 2 \cdot 1 - 2 i

Factor out common term

2

= 2 (1 - i)

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

Cancel the common factor:

2

= \frac{1 - i}{2}

Rewrite

\frac{1 - i}{2}

in standard complex form:

\frac{1}{2} - \frac{1}{2} i

\frac{1 - i}{2}

Apply the fraction rule:

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

\frac{1 - i}{2} = \frac{1}{2} - \frac{i}{2}

= \frac{1}{2} - \frac{i}{2}

= \frac{1}{2} - \frac{1}{2} i

The solutions to the quadratic equation are:

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

The solutions are

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

Substitute back

u = sin (x)

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

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

No Solution

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

No Solution

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

No Solution

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

No Solution

Combine all the solutions

x = \frac{3 π}{2} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for cos(2x)sin(x)=1 ?
The general solution for cos(2x)sin(x)=1 is x=(3pi)/2+2pin