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

The general solution for sin(2x)-2cos(x)=0 is x= pi/2+2pin,x=(3pi)/2+2pin

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

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

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

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

Rewrite using trig identities

sin (2 x) - 2 cos (x)

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

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

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

Factor

- 2 cos (x) + 2 cos (x) sin (x) : 2 cos (x) (sin (x) - 1)

- 2 cos (x) + 2 cos (x) sin (x)

Factor out common term

2 cos (x)

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

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

Solving each part separately

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

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

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^{2} (θ) - cos^{2} (θ) = 1 + cos (θ), 0 \leq θ < 2 π

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

sin^{2} (θ) - sin (θ) = 0

2 cos^{2} (x) - 1 = 0, 0 \leq x < 2 π

cos (x + \frac{π}{2}) = sin (x)

What is the general solution for sin(2x)-2cos(x)=0 ?
The general solution for sin(2x)-2cos(x)=0 is x= pi/2+2pin,x=(3pi)/2+2pin