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

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

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

Solution steps

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

Subtract

(sin (x))^{2}

from both sides

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

Rewrite using trig identities

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

Solve by substitution

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} :

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

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

\frac{tan ( 2 θ )}{2} = tan (θ)

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

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

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

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