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

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

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

2 cos (x) + sin (2 x) = 0, 0 \leq x \leq 2 π

x = \frac{π}{2}, x = \frac{3 π}{2}

Degrees

x = 9 0^{\circ}, x = 27 0^{\circ}

Solution steps

2 cos (x) + sin (2 x) = 0, 0 \leq x \leq 2 π

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, 0 \leq x \leq 2 π : x = \frac{π}{2}, x = \frac{3 π}{2}

cos (x) = 0, 0 \leq x \leq 2 π

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

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{2}, x = \frac{3 π}{2}

sin (x) + 1 = 0, 0 \leq x \leq 2 π : x = \frac{3 π}{2}

sin (x) + 1 = 0, 0 \leq x \leq 2 π

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

Solutions for the range

0 \leq x \leq 2 π

x = \frac{3 π}{2}

Combine all the solutions

x = \frac{π}{2}, x = \frac{3 π}{2}

tan^{2} (x) sin (x) = sin (x)

2 sin (2 x) - 3 = 0

2 sin (x) cos (x) = 0

tan^{5} (x) - 9 tan (x) = 0

1 - cos (θ) = \frac{1}{2}

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