What is the general solution for cos(2x)-sin(x)=1,0<= x<2pi ?

Q: What is the general solution for cos(2x)-sin(x)=1,0<= x<2pi ?

The general solution for cos(2x)-sin(x)=1,0<= x<2pi is x=(7pi)/6 ,x=(11pi)/6 ,x=0,x=pi

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cos(2x)-sin(x)=1,0<= x<2pi

Solution

cos (2 x) - sin (x) = 1, 0 \leq x < 2 π

Solution

x = \frac{7 π}{6}, x = \frac{11 π}{6}, x = 0, x = π

Degrees

x = 21 0^{\circ}, x = 33 0^{\circ}, x = 0^{\circ}, x = 18 0^{\circ}

Solution steps

cos (2 x) - sin (x) = 1, 0 \leq x < 2 π

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)

Simplify

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

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

Solve by substitution

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

Let:

sin (x) = u

- u - 2 u^{2} = 0

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

- u - 2 u^{2} = 0

Write in the standard form

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

- 2 u^{2} - u = 0

Solve with the quadratic formula

- 2 u^{2} - u = 0

Quadratic Equation Formula:

For

a = - 2, b = - 1, c = 0

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 2 ) \cdot 0}{2 ( - 2 )}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 2 ) \cdot 0}{2 ( - 2 )}

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

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

Apply rule

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 2 \cdot 0 = 0

4 \cdot 2 \cdot 0

Apply rule

0 \cdot a = 0

= 0

= 1 + 0

Add the numbers:

1 + 0 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

\frac{- ( - 1 ) + 1}{2 ( - 2 )}

Remove parentheses:

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

= \frac{1 + 1}{- 2 \cdot 2}

Add the numbers:

1 + 1 = 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}

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

\frac{- ( - 1 ) - 1}{2 ( - 2 )}

Remove parentheses:

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

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

Subtract the numbers:

1 - 1 = 0

= \frac{0}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{0}{- 4}

Apply the fraction rule:

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

= - \frac{0}{4}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

The solutions to the quadratic equation are:

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

Substitute back

u = sin (x)

sin (x) = - \frac{1}{2}, sin (x) = 0

sin (x) = - \frac{1}{2}, sin (x) = 0

sin (x) = - \frac{1}{2}, 0 \leq x < 2 π : x = \frac{7 π}{6}, x = \frac{11 π}{6}

sin (x) = - \frac{1}{2}, 0 \leq x < 2 π

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{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Solutions for the range

0 \leq x < 2 π

x = \frac{7 π}{6}, x = \frac{11 π}{6}

sin (x) = 0, 0 \leq x < 2 π : x = 0, x = π

sin (x) = 0, 0 \leq x < 2 π

General solutions for

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

Solutions for the range

0 \leq x < 2 π

x = 0, x = π

Combine all the solutions

x = \frac{7 π}{6}, x = \frac{11 π}{6}, x = 0, x = π

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Frequently Asked Questions (FAQ)

What is the general solution for cos(2x)-sin(x)=1,0<= x<2pi ?
The general solution for cos(2x)-sin(x)=1,0<= x<2pi is x=(7pi)/6 ,x=(11pi)/6 ,x=0,x=pi