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

The general solution for sin^2(x)=1-cos(x),0<= x<= 2pi is x= pi/2 ,x=(3pi)/2 ,x=0,x=2pi

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

sin^{2} (x) = 1 - cos (x), 0 \leq x \leq 2 π

x = \frac{π}{2}, x = \frac{3 π}{2}, x = 0, x = 2 π

Solution steps

sin^{2} (x) = 1 - cos (x), 0 \leq x \leq 2 π

Subtract

1 - cos (x)

from both sides

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

Rewrite using trig identities

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

Solve by substitution

cos (x) = 0, cos (x) = 1

cos (x) = 0, 0 \leq x \leq 2 π : x = \frac{π}{2}, x = \frac{3 π}{2}

cos (x) = 1, 0 \leq x \leq 2 π : x = 0, x = 2 π

Combine all the solutions

x = \frac{π}{2}, x = \frac{3 π}{2}, x = 0, x = 2 π

cos (\frac{π}{4} c) = sin (2 (\frac{π}{4}))

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

sin (4 θ) = 0.345

- 3 = cot^{2} (x) + 3 csc (x)

4 cos^{2} (θ) + 3 cos (θ) = 1

What is the general solution for sin^2(x)=1-cos(x),0<= x<= 2pi ?
The general solution for sin^2(x)=1-cos(x),0<= x<= 2pi is x= pi/2 ,x=(3pi)/2 ,x=0,x=2pi