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

The general solution for sin^2(x)-2cos(x)=0 is x=1.14371…+2pin,x=2pi-1.14371…+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

x = 1.14371 \dots + 2 πn, x = 2 π - 1.14371 \dots + 2 πn

Degrees

x = 65.53019 \dots^{\circ} + 36 0^{\circ} n, x = 294.46980 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

1 - cos^{2} (x) - 2 cos (x) = 0

Solve by substitution

1 - cos^{2} (x) - 2 cos (x) = 0

Let:

cos (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

(- 2)^{2} - 4 (- 1) \cdot 1 = 22

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} + 4

2^{2} = 4

= 4 + 4

Add the numbers:

4 + 4 = 8

= 8

Prime factorization of

8 : 2^{3}

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

is a prime number, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

Apply exponent rule:

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

= 2^{2} \cdot 2

Apply radical rule:

= 2 2^{2}

Apply radical rule:

2^{2} = 2

= 22

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

Separate the solutions

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

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

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

Remove parentheses:

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

Apply the fraction rule:

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

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

Cancel

\frac{2 + 2 2}{2} : 1 + 2

\frac{2 + 2 2}{2}

Factor

2 + 22 : 2 (1 + 2)

2 + 22

Rewrite as

= 2 \cdot 1 + 22

Factor out common term

2

= 2 (1 + 2)

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

Divide the numbers:

\frac{2}{2} = 1

= 1 + 2

= - (1 + 2)

Distribute parentheses

= - (1) - (2)

Apply minus-plus rules

+ (- a) = - a

= - 1 - 2

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

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

Remove parentheses:

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

Apply the fraction rule:

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

2 - 22 = - (22 - 2)

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

Factor

22 - 2 : 2 (2 - 1)

22 - 2

Rewrite as

= 22 - 2 \cdot 1

Factor out common term

2

= 2 (2 - 1)

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

Divide the numbers:

\frac{2}{2} = 1

= 2 - 1

The solutions to the quadratic equation are:

u = - 1 - 2, u = 2 - 1

Substitute back

u = cos (x)

cos (x) = - 1 - 2, cos (x) = 2 - 1

cos (x) = - 1 - 2, cos (x) = 2 - 1

cos (x) = - 1 - 2 :

No Solution

cos (x) = - 1 - 2

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = 2 - 1 : x = arccos (2 - 1) + 2 πn, x = 2 π - arccos (2 - 1) + 2 πn

cos (x) = 2 - 1

Apply trig inverse properties

cos (x) = 2 - 1

General solutions for

cos (x) = 2 - 1

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (2 - 1) + 2 πn, x = 2 π - arccos (2 - 1) + 2 πn

x = arccos (2 - 1) + 2 πn, x = 2 π - arccos (2 - 1) + 2 πn

Combine all the solutions

x = arccos (2 - 1) + 2 πn, x = 2 π - arccos (2 - 1) + 2 πn

Show solutions in decimal form

x = 1.14371 \dots + 2 πn, x = 2 π - 1.14371 \dots + 2 πn

Popular Examples

3sin^2(θ)+3sin(θ)-6=-4sin(θ)

3 sin^{2} (θ) + 3 sin (θ) - 6 = - 4 sin (θ)

sin(4x)-cos(3x)=sin(2x)

sin (4 x) - cos (3 x) = sin (2 x)

sin(x)cos(x)= 1/2 tan(x)

sin (x) cos (x) = \frac{1}{2} tan (x)

solvefor x,2sec^2(x)+tan^2(x)-3=0 solve for

x, 2 sec^{2} (x) + tan^{2} (x) - 3 = 0

5sin(θ/2)=5cos(θ/2)

5 sin (\frac{θ}{2}) = 5 cos (\frac{θ}{2})

Frequently Asked Questions (FAQ)

What is the general solution for sin^2(x)-2cos(x)=0 ?
The general solution for sin^2(x)-2cos(x)=0 is x=1.14371…+2pin,x=2pi-1.14371…+2pin