What is the general solution for sin(x)cos(x)=-9cos(x) ?

The general solution for sin(x)cos(x)=-9cos(x) is x= pi/2+2pin,x=(3pi)/2+2pin

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sin(x)cos(x)=-9cos(x)

sin (x) cos (x) = - 9 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

sin (x) cos (x) = - 9 cos (x)

Subtract

- 9 cos (x)

from both sides

sin (x) cos (x) + 9 cos (x) = 0

Factor

sin (x) cos (x) + 9 cos (x) : cos (x) (sin (x) + 9)

sin (x) cos (x) + 9 cos (x)

Factor out common term

cos (x)

= cos (x) (sin (x) + 9)

cos (x) (sin (x) + 9) = 0

Solving each part separately

cos (x) = 0 or sin (x) + 9 = 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) + 9 = 0 :

No Solution

sin (x) + 9 = 0

Move

9

to the right side

sin (x) + 9 = 0

Subtract

9

from both sides

sin (x) + 9 - 9 = 0 - 9

Simplify

sin (x) = - 9

sin (x) = - 9

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

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

5 tan (θ) - 7 = 0

cot^{2} (x) + 1 = sin^{2} (x)

(4 + tan (x)) (3 sin (x) - 2) = 0

sec (2 x) - 1 = 0

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

What is the general solution for sin(x)cos(x)=-9cos(x) ?
The general solution for sin(x)cos(x)=-9cos(x) is x= pi/2+2pin,x=(3pi)/2+2pin