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

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

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

solve for

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

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

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

Switch sides

- 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{3 π}{2} + 2 πn

sin (x) + 1 = 0

Move

1

to the right side

sin (x) + 1 = 0

Subtract

1

from 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{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

sec (θ) = \frac{5}{2}

4 cos (x - 1) = 0

cos (θ) = \frac{5}{7}, cos (\frac{θ}{2}), 0^{\circ} < θ < 9 0^{\circ}

sin (x) + 4 = cos (x) + 3

\frac{sin ( 17 0 ^{\circ} )}{8} = \frac{sin ( B )}{5}

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