What is the general solution for (2sin(x)-1)(cos(x)-sqrt(2))=0 ?

The general solution for (2sin(x)-1)(cos(x)-sqrt(2))=0 is x= pi/6+2pin,x=(5pi)/6+2pin

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

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

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Solving each part separately

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

2 sin (x) - 1 = 0 : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

2 sin (x) - 1 = 0

Move

1

to the right side

2 sin (x) - 1 = 0

Add

1

to both sides

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

Simplify

2 sin (x) = 1

2 sin (x) = 1

Divide both sides by

2

2 sin (x) = 1

Divide both sides by

2

\frac{2 sin ( x )}{2} = \frac{1}{2}

Simplify

sin (x) = \frac{1}{2}

sin (x) = \frac{1}{2}

General solutions for

sin (x) = \frac{1}{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{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

cos (x) - 2 = 0 :

No Solution

cos (x) - 2 = 0

Move

2

to the right side

cos (x) - 2 = 0

Add

2

to both sides

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

Simplify

cos (x) = 2

cos (x) = 2

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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}

sin (x) = \frac{10}{300}

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

What is the general solution for (2sin(x)-1)(cos(x)-sqrt(2))=0 ?
The general solution for (2sin(x)-1)(cos(x)-sqrt(2))=0 is x= pi/6+2pin,x=(5pi)/6+2pin