What is the general solution for 2sin(x)+sqrt(2)=0,0<= x<= 2pi ?

The general solution for 2sin(x)+sqrt(2)=0,0<= x<= 2pi is x=(5pi)/4 ,x=(7pi)/4

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2sin(x)+sqrt(2)=0,0<= x<= 2pi

2 sin (x) + 2 = 0, 0 \leq x \leq 2 π

x = \frac{5 π}{4}, x = \frac{7 π}{4}

Degrees

x = 22 5^{\circ}, x = 31 5^{\circ}

Solution steps

2 sin (x) + 2 = 0, 0 \leq x \leq 2 π

Move

2

to the right side

2 sin (x) + 2 = 0

Subtract

2

from both sides

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

Simplify

2 sin (x) = - 2

2 sin (x) = - 2

Divide both sides by

2

2 sin (x) = - 2

Divide both sides by

2

\frac{2 sin ( x )}{2} = \frac{- 2}{2}

Simplify

sin (x) = - \frac{2}{2}

sin (x) = - \frac{2}{2}

General solutions for

sin (x) = - \frac{2}{2}

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{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = \frac{5 π}{4}, x = \frac{7 π}{4}

2 cos^{2} (x) + 3 cos (x) = 2, 0 \leq x \leq 2 π

cos (θ) = \frac{7}{11}

(1 - sin^{2} (x)) \cdot 2 - 3 sin (x) = 0

sin (\frac{θ}{2}) = sin (θ)

2 tan (x) = 1

What is the general solution for 2sin(x)+sqrt(2)=0,0<= x<= 2pi ?
The general solution for 2sin(x)+sqrt(2)=0,0<= x<= 2pi is x=(5pi)/4 ,x=(7pi)/4