What is the general solution for -2cos^2(x)-5sin(x)+5=0 ?

The general solution for -2cos^2(x)-5sin(x)+5=0 is x= pi/2+2pin

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Popular >

-2cos^2(x)-5sin(x)+5=0

Solution

- 2 cos^{2} (x) - 5 sin (x) + 5 = 0

Solution

x = \frac{π}{2} + 2 πn

Degrees

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

Solution steps

- 2 cos^{2} (x) - 5 sin (x) + 5 = 0

Rewrite using trig identities

5 - 2 cos^{2} (x) - 5 sin (x)

Use the Pythagorean identity:

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

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

= 5 - 2 (1 - sin^{2} (x)) - 5 sin (x)

Simplify

5 - 2 (1 - sin^{2} (x)) - 5 sin (x) : 2 sin^{2} (x) - 5 sin (x) + 3

5 - 2 (1 - sin^{2} (x)) - 5 sin (x)

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

a = - 2, b = 1, c = sin^{2} (x)

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

Apply minus-plus rules

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

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

= 5 - 2 + 2 sin^{2} (x) - 5 sin (x)

Subtract the numbers:

5 - 2 = 3

= 2 sin^{2} (x) - 5 sin (x) + 3

= 2 sin^{2} (x) - 5 sin (x) + 3

3 + 2 sin^{2} (x) - 5 sin (x) = 0

Solve by substitution

3 + 2 sin^{2} (x) - 5 sin (x) = 0

Let:

sin (x) = u

3 + 2 u^{2} - 5 u = 0

3 + 2 u^{2} - 5 u = 0 : u = \frac{3}{2}, u = 1

3 + 2 u^{2} - 5 u = 0

Write in the standard form

a x^{2} + b x + c = 0

2 u^{2} - 5 u + 3 = 0

Solve with the quadratic formula

2 u^{2} - 5 u + 3 = 0

Quadratic Equation Formula:

For

a = 2, b = - 5, c = 3

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

(- 5)^{2} - 4 \cdot 2 \cdot 3 = 1

(- 5)^{2} - 4 \cdot 2 \cdot 3

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 5)^{2} = 5^{2}

= 5^{2} - 4 \cdot 2 \cdot 3

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 5^{2} - 24

5^{2} = 25

= 25 - 24

Subtract the numbers:

25 - 24 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- ( - 5 ) \pm 1}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 1}{2 \cdot 2}, u_{2} = \frac{- ( - 5 ) - 1}{2 \cdot 2}

u = \frac{- ( - 5 ) + 1}{2 \cdot 2} : \frac{3}{2}

\frac{- ( - 5 ) + 1}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{5 + 1}{2 \cdot 2}

Add the numbers:

5 + 1 = 6

= \frac{6}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{6}{4}

Cancel the common factor:

2

= \frac{3}{2}

u = \frac{- ( - 5 ) - 1}{2 \cdot 2} : 1

\frac{- ( - 5 ) - 1}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{5 - 1}{2 \cdot 2}

Subtract the numbers:

5 - 1 = 4

= \frac{4}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = \frac{3}{2}, u = 1

Substitute back

u = sin (x)

sin (x) = \frac{3}{2}, sin (x) = 1

sin (x) = \frac{3}{2}, sin (x) = 1

sin (x) = \frac{3}{2} :

No Solution

sin (x) = \frac{3}{2}

- 1 \leq sin (x) \leq 1

No Solution

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

Combine all the solutions

x = \frac{π}{2} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for -2cos^2(x)-5sin(x)+5=0 ?
The general solution for -2cos^2(x)-5sin(x)+5=0 is x= pi/2+2pin