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

The general solution for -sin(x)(2+sin(x))-cos^2(x)=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin

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

Solution

- sin (x) (2 + sin (x)) - cos^{2} (x) = 0

Solution

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

Degrees

x = 21 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n

Solution steps

- sin (x) (2 + sin (x)) - cos^{2} (x) = 0

Rewrite using trig identities

- cos^{2} (x) - (2 + sin (x)) sin (x)

Use the Pythagorean identity:

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

cos^{2} (x) = 1 - sin^{2} (x)

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

Simplify

- (1 - sin^{2} (x)) - (2 + sin (x)) sin (x) : - 2 sin (x) - 1

- (1 - sin^{2} (x)) - (2 + sin (x)) sin (x)

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

- (1 - sin^{2} (x)) : - 1 + sin^{2} (x)

- (1 - sin^{2} (x))

Distribute parentheses

= - (1) - (- sin^{2} (x))

Apply minus-plus rules

- (- a) = a, - (a) = - a

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

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

Expand

- sin (x) (2 + sin (x)) : - 2 sin (x) - sin^{2} (x)

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

Apply the distributive law:

a (b + c) = ab + a c

a = - sin (x), b = 2, c = sin (x)

= - sin (x) \cdot 2 + (- sin (x)) sin (x)

Apply minus-plus rules

+ (- a) = - a

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

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

sin (x) sin (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

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

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

Simplify

- 1 + sin^{2} (x) - 2 sin (x) - sin^{2} (x) : - 2 sin (x) - 1

- 1 + sin^{2} (x) - 2 sin (x) - sin^{2} (x)

Group like terms

= sin^{2} (x) - 2 sin (x) - sin^{2} (x) - 1

Add similar elements:

sin^{2} (x) - sin^{2} (x) = 0

= - 2 sin (x) - 1

= - 2 sin (x) - 1

= - 2 sin (x) - 1

- 1 - 2 sin (x) = 0

Move

1

to the right side

- 1 - 2 sin (x) = 0

Add

1

to both sides

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

Simplify

- 2 sin (x) = 1

- 2 sin (x) = 1

Divide both sides by

- 2

- 2 sin (x) = 1

Divide both sides by

- 2

\frac{- 2 sin ( x )}{- 2} = \frac{1}{- 2}

Simplify

sin (x) = - \frac{1}{2}

sin (x) = - \frac{1}{2}

General solutions for

sin (x) = - \frac{1}{2}

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

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

Graph

Popular Examples

cos(b)=0.3125

cos (b) = 0.3125

sin(1/2 x)-1=0

sin (\frac{1}{2} x) - 1 = 0

-1/16 sin(x/4)=0

- \frac{1}{16} sin (\frac{x}{4}) = 0

sec(x)=-2,pi<x<(3pi)/2

sec (x) = - 2, π < x < \frac{3 π}{2}

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

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

Frequently Asked Questions (FAQ)

What is the general solution for -sin(x)(2+sin(x))-cos^2(x)=0 ?
The general solution for -sin(x)(2+sin(x))-cos^2(x)=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin