What is the general solution for 1+cos(x)=sqrt(3)-sin(x) ?

The general solution for 1+cos(x)=sqrt(3)-sin(x) is x=0.54408…+2pin-pi/4 ,x=pi-0.54408…+2pin-pi/4

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1+cos(x)=sqrt(3)-sin(x)

1 + cos (x) = 3 - sin (x)

x = 0.54408 \dots + 2 πn - \frac{π}{4}, x = π - 0.54408 \dots + 2 πn - \frac{π}{4}

Solution steps

1 + cos (x) = 3 - sin (x)

Subtract

3 - sin (x)

from both sides

1 + cos (x) - 3 + sin (x) = 0

Rewrite using trig identities

1 - 3 + 2 sin (x + \frac{π}{4}) = 0

Move

1

to the right side

- 3 + 2 sin (x + \frac{π}{4}) = - 1

Move

3

to the right side

2 sin (x + \frac{π}{4}) = - 1 + 3

Divide both sides by

2

sin (x + \frac{π}{4}) = \frac{2 ( - 1 + 3 )}{2}

Apply trig inverse properties

x + \frac{π}{4} = arcsin (\frac{2 ( - 1 + 3 )}{2}) + 2 πn, x + \frac{π}{4} = π - arcsin (\frac{2 ( - 1 + 3 )}{2}) + 2 πn

Solve

x + \frac{π}{4} = arcsin (\frac{2 ( - 1 + 3 )}{2}) + 2 πn : x = arcsin (\frac{- 1 + 3}{2}) + 2 πn - \frac{π}{4}

Solve

x + \frac{π}{4} = π - arcsin (\frac{2 ( - 1 + 3 )}{2}) + 2 πn : x = π - arcsin (\frac{- 1 + 3}{2}) + 2 πn - \frac{π}{4}

x = arcsin (\frac{- 1 + 3}{2}) + 2 πn - \frac{π}{4}, x = π - arcsin (\frac{- 1 + 3}{2}) + 2 πn - \frac{π}{4}

Show solutions in decimal form

x = 0.54408 \dots + 2 πn - \frac{π}{4}, x = π - 0.54408 \dots + 2 πn - \frac{π}{4}

3 = 7 cos (\frac{π}{3} t)

cos^{2} (2 x) = cos (4 x)

θ, tan (2 θ) = - 1

3 cos (z + 1.2) = 2

\frac{1}{2} = \frac{3}{2} \cdot cos (x)

What is the general solution for 1+cos(x)=sqrt(3)-sin(x) ?
The general solution for 1+cos(x)=sqrt(3)-sin(x) is x=0.54408…+2pin-pi/4 ,x=pi-0.54408…+2pin-pi/4