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

The general solution for (sin(x)+cos(x))^2=1+sin^2(x) is x=2pin,x=pi+2pin,x=1.10714…+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

x = 2 πn, x = π + 2 πn, x = 1.10714 \dots + πn

Solution steps

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

Subtract

1 + sin^{2} (x)

from both sides

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

Rewrite using trig identities

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

Factor

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

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

Solving each part separately

sin (x) = 0 or - sin (x) + 2 cos (x) = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

- sin (x) + 2 cos (x) = 0 : x = arctan (2) + πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = arctan (2) + πn

Show solutions in decimal form

x = 2 πn, x = π + 2 πn, x = 1.10714 \dots + πn

x, y = sin (x) y

1 \cdot sin (x) = 1.5 \cdot sin (\frac{x}{2})

sin (3 θ) = - 1, 0 \leq x \leq 2 π

\frac{sin ( 9 0 ^{\circ} )}{5} = \frac{sin ( x )}{4}

2 sin^{2} (x) = - 5 sin (x) - 3

What is the general solution for (sin(x)+cos(x))^2=1+sin^2(x) ?
The general solution for (sin(x)+cos(x))^2=1+sin^2(x) is x=2pin,x=pi+2pin,x=1.10714…+pin