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cos(x)(2sin(x)+1)=0,0<= x<= 2pi

Solution

cos (x) (2 sin (x) + 1) = 0, 0 \leq x \leq 2 π

x = \frac{π}{2}, x = \frac{3 π}{2}, x = \frac{7 π}{6}, x = \frac{11 π}{6}

Degrees

x = 9 0^{\circ}, x = 27 0^{\circ}, x = 21 0^{\circ}, x = 33 0^{\circ}

Solution steps

cos (x) (2 sin (x) + 1) = 0, 0 \leq x \leq 2 π

Solving each part separately

cos (x) = 0 or 2 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}

2 sin (x) + 1 = 0, 0 \leq x \leq 2 π : x = \frac{7 π}{6}, x = \frac{11 π}{6}

2 sin (x) + 1 = 0, 0 \leq x \leq 2 π

Move

1

to the right side

2 sin (x) + 1 = 0

Subtract

1

from 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