What is the general solution for sin(piy)=cos(1)sin(pi/6)-1/2 ?

The general solution for sin(piy)=cos(1)sin(pi/6)-1/2 is y=(-0.23192…)/pi+2n,y=1+(0.23192…)/pi+2n

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin(piy)=cos(1)sin(pi/6)-1/2

sin (π y) = cos (1) sin (\frac{π}{6}) - \frac{1}{2}

y = \frac{- 0.23192 \dots}{π} + 2 n, y = 1 + \frac{0.23192 \dots}{π} + 2 n

Solution steps

sin (π y) = cos (1) sin (\frac{π}{6}) - \frac{1}{2}

sin (\frac{π}{6}) = \frac{1}{2}

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

Apply trig inverse properties

π y = arcsin (cos (1) \frac{1}{2} - \frac{1}{2}) + 2 πn, π y = π + arcsin (- cos (1) \frac{1}{2} + \frac{1}{2}) + 2 πn

Solve

π y = arcsin (cos (1) \frac{1}{2} - \frac{1}{2}) + 2 πn : y = \frac{arcsin ( \frac{1}{2} cos ( 1 ) - \frac{1}{2} )}{π} + 2 n

Solve

π y = π + arcsin (- cos (1) \frac{1}{2} + \frac{1}{2}) + 2 πn : y = 1 + \frac{arcsin ( - \frac{1}{2} cos ( 1 ) + \frac{1}{2} )}{π} + 2 n

y = \frac{arcsin ( \frac{1}{2} cos ( 1 ) - \frac{1}{2} )}{π} + 2 n, y = 1 + \frac{arcsin ( - \frac{1}{2} cos ( 1 ) + \frac{1}{2} )}{π} + 2 n

Show solutions in decimal form

y = \frac{- 0.23192 \dots}{π} + 2 n, y = 1 + \frac{0.23192 \dots}{π} + 2 n

0 = 4 cos (120 x)

(2 sin (x) - 1) (cos (x) - 2) = 0

7 - 6 sin (x) = 3 + 2 sin (x)

cos (θ) = \frac{2 \cdot 1 0 ^{- 5}}{54 \cdot 1 0 ^{- 6}}

sin (a) = \frac{9}{41}

What is the general solution for sin(piy)=cos(1)sin(pi/6)-1/2 ?
The general solution for sin(piy)=cos(1)sin(pi/6)-1/2 is y=(-0.23192…)/pi+2n,y=1+(0.23192…)/pi+2n