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

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

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

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

1 - sin^{2} (x)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Factor

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

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

Factor out common term

cos^{2} (x)

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

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

Solving each part separately

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

cos^{2} (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos^{2} (x) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

cos (x) = 0

General solutions for

cos (x) = 0

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

sin (x) - 1 = 0 : x = \frac{π}{2} + 2 πn

sin (x) - 1 = 0

Move

1

to the right side

sin (x) - 1 = 0

Add

1

to both sides

sin (x) - 1 + 1 = 0 + 1

Simplify

sin (x) = 1

sin (x) = 1

General solutions for

sin (x) = 1

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

sin (θ) = 1 - (1 - (cos (θ) - 1)^{2})

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

14.79 m = (\frac{7.6 1 _{0}^{2} sin ( 2 θ _{0} )}{g})

2 cos (x) + 3 cos (x) - 2 = 0

sec (\frac{5 θ}{4}) = 2, [0.2 π]

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