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

The general solution for cos(x)=(sin(x))^2 is x=0.90455…+2pin,x=2pi-0.90455…+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

x = 0.90455 \dots + 2 πn, x = 2 π - 0.90455 \dots + 2 πn

Degrees

x = 51.82729 \dots^{\circ} + 36 0^{\circ} n, x = 308.17270 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

(sin (x))^{2}

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

- 1 + u + u^{2} = 0 : u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

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

Write in the standard form

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

u^{2} + u - 1 = 0

Solve with the quadratic formula

u^{2} + u - 1 = 0

Quadratic Equation Formula:

For

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

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

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

1^{2} - 4 \cdot 1 \cdot (- 1) = 5

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 1)

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

Add the numbers:

1 + 4 = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 1 + 5}{2 \cdot 1}, u_{2} = \frac{- 1 - 5}{2 \cdot 1}

u = \frac{- 1 + 5}{2 \cdot 1} : \frac{- 1 + 5}{2}

\frac{- 1 + 5}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 + 5}{2}

u = \frac{- 1 - 5}{2 \cdot 1} : \frac{- 1 - 5}{2}

\frac{- 1 - 5}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 - 5}{2}

The solutions to the quadratic equation are:

u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

Substitute back

u = cos (x)

cos (x) = \frac{- 1 + 5}{2}, cos (x) = \frac{- 1 - 5}{2}

cos (x) = \frac{- 1 + 5}{2}, cos (x) = \frac{- 1 - 5}{2}

cos (x) = \frac{- 1 + 5}{2} : x = arccos (\frac{- 1 + 5}{2}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

cos (x) = \frac{- 1 + 5}{2}

Apply trig inverse properties

cos (x) = \frac{- 1 + 5}{2}

General solutions for

cos (x) = \frac{- 1 + 5}{2}

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

x = arccos (\frac{- 1 + 5}{2}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

x = arccos (\frac{- 1 + 5}{2}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

cos (x) = \frac{- 1 - 5}{2} :

No Solution

cos (x) = \frac{- 1 - 5}{2}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = arccos (\frac{- 1 + 5}{2}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

Show solutions in decimal form

x = 0.90455 \dots + 2 πn, x = 2 π - 0.90455 \dots + 2 πn

Popular Examples

(20)/(sin(35))=(16)/(sin(C))

\frac{20}{sin ( 3 5 ^{\circ} )} = \frac{16}{sin ( C )}

(tan(2θ))/2 =tan(θ)

\frac{tan ( 2 θ )}{2} = tan (θ)

2sin^2(3x)-3cos(3x)+1=0,0<= x<= 180

2 sin^{2} (3 x) - 3 cos (3 x) + 1 = 0, 0^{\circ} \leq x \leq 18 0^{\circ}

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

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

sin(x)=(-sin(2x))/2

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

Frequently Asked Questions (FAQ)

What is the general solution for cos(x)=(sin(x))^2 ?
The general solution for cos(x)=(sin(x))^2 is x=0.90455…+2pin,x=2pi-0.90455…+2pin