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Popular Trigonometry >

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

Solution

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

Solution

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

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 21 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Expand

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

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

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

(- 1 - sin (x) + 2 sin^{2} (x)) \cdot 2 = 0

Solve by substitution

(- 1 - sin (x) + 2 sin^{2} (x)) \cdot 2 = 0

Let:

sin (x) = u

(- 1 - u + 2 u^{2}) \cdot 2 = 0

(- 1 - u + 2 u^{2}) \cdot 2 = 0 : u = 1, u = - \frac{1}{2}

(- 1 - u + 2 u^{2}) \cdot 2 = 0

Factor

(- 1 - u + 2 u^{2}) \cdot 2 : 2 (u - 1) (2 u + 1)

(- 1 - u + 2 u^{2}) \cdot 2

Factor

2 u^{2} - u - 1 : (2 u + 1) (u - 1)

2 u^{2} - u - 1

Write in the standard form

a x^{2} + b x + c

= 2 u^{2} - u - 1

Break the expression into groups

2 u^{2} - u - 1

Definition

Factors of

2 : 1, 2

2

Divisors (Factors)

Find the Prime factors of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Add 1

1

The factors of

2

1, 2

Negative factors of

2 : - 1, - 2

Multiply the factors by

- 1

to get the negative factors

- 1, - 2

For every two factors such that

u * v = - 2,

check if

u + v = - 1

Check

u = 1, v = - 2 : u * v = - 2, u + v = - 1 \Rightarrow

TrueCheck

u = 2, v = - 1 : u * v = - 2, u + v = 1 \Rightarrow

False

u = 1, v = - 2

Group into

(a x^{2} + ux) + (vx + c)

(2 u^{2} + u) + (- 2 u - 1)

= (2 u^{2} + u) + (- 2 u - 1)

Factor out

u

from

2 u^{2} + u : u (2 u + 1)

2 u^{2} + u

Apply exponent rule:

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

u^{2} = uu

= 2 uu + u

Factor out common term

u

= u (2 u + 1)

Factor out

- 1

from

- 2 u - 1 : - (2 u + 1)

- 2 u - 1

Factor out common term

- 1

= - (2 u + 1)

= u (2 u + 1) - (2 u + 1)

Factor out common term

2 u + 1

= (2 u + 1) (u - 1)

= 2 (u - 1) (2 u + 1)

2 (u - 1) (2 u + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u - 1 = 0 or 2 u + 1 = 0

Solve

u - 1 = 0 : u = 1

u - 1 = 0

Move

1

to the right side

u - 1 = 0

Add

1

to both sides

u - 1 + 1 = 0 + 1

Simplify

u = 1

u = 1

Solve

2 u + 1 = 0 : u = - \frac{1}{2}

2 u + 1 = 0

Move

1

to the right side

2 u + 1 = 0

Subtract

1

from both sides

2 u + 1 - 1 = 0 - 1

Simplify

2 u = - 1

2 u = - 1

Divide both sides by

2

2 u = - 1

Divide both sides by

2

\frac{2 u}{2} = \frac{- 1}{2}

Simplify

u = - \frac{1}{2}

u = - \frac{1}{2}

The solutions to the quadratic equation are:

u = 1, u = - \frac{1}{2}

Substitute back

u = sin (x)

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

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

sin (x) = 1 : x = \frac{π}{2} + 2 πn

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{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

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

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

Combine all the solutions

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

Graph

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