What is the general solution for 2cos^2(x)-9sin(x)-3=0 ?

The general solution for 2cos^2(x)-9sin(x)-3=0 is x=-0.11424…+2pin,x=pi+0.11424…+2pin

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Popular >

2cos^2(x)-9sin(x)-3=0

Solution

2 cos^{2} (x) - 9 sin (x) - 3 = 0

Solution

x = - 0.11424 \dots + 2 πn, x = π + 0.11424 \dots + 2 πn

Degrees

x = - 6.54589 \dots^{\circ} + 36 0^{\circ} n, x = 186.54589 \dots^{\circ} + 36 0^{\circ} n

Solution steps

2 cos^{2} (x) - 9 sin (x) - 3 = 0

Rewrite using trig identities

- 3 + 2 cos^{2} (x) - 9 sin (x)

Use the Pythagorean identity:

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

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

= - 3 + 2 (1 - sin^{2} (x)) - 9 sin (x)

Simplify

- 3 + 2 (1 - sin^{2} (x)) - 9 sin (x) : - 2 sin^{2} (x) - 9 sin (x) - 1

- 3 + 2 (1 - sin^{2} (x)) - 9 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)

Multiply the numbers:

2 \cdot 1 = 2

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

= - 3 + 2 - 2 sin^{2} (x) - 9 sin (x)

Add/Subtract the numbers:

- 3 + 2 = - 1

= - 2 sin^{2} (x) - 9 sin (x) - 1

= - 2 sin^{2} (x) - 9 sin (x) - 1

- 1 - 2 sin^{2} (x) - 9 sin (x) = 0

Solve by substitution

- 1 - 2 sin^{2} (x) - 9 sin (x) = 0

Let:

sin (x) = u

- 1 - 2 u^{2} - 9 u = 0

- 1 - 2 u^{2} - 9 u = 0 : u = - \frac{9 + 73}{4}, u = - \frac{9 - 73}{4}

- 1 - 2 u^{2} - 9 u = 0

Write in the standard form

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

- 2 u^{2} - 9 u - 1 = 0

Solve with the quadratic formula

- 2 u^{2} - 9 u - 1 = 0

Quadratic Equation Formula:

For

a = - 2, b = - 9, c = - 1

u_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 ( - 2 ) ( - 1 )}{2 ( - 2 )}

u_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 ( - 2 ) ( - 1 )}{2 ( - 2 )}

(- 9)^{2} - 4 (- 2) (- 1) = 73

(- 9)^{2} - 4 (- 2) (- 1)

Apply rule

- (- a) = a

= (- 9)^{2} - 4 \cdot 2 \cdot 1

Apply exponent rule:

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

n

is even

(- 9)^{2} = 9^{2}

= 9^{2} - 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 9^{2} - 8

9^{2} = 81

= 81 - 8

Subtract the numbers:

81 - 8 = 73

= 73

u_{1, 2} = \frac{- ( - 9 ) \pm 73}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- ( - 9 ) + 73}{2 ( - 2 )}, u_{2} = \frac{- ( - 9 ) - 73}{2 ( - 2 )}

u = \frac{- ( - 9 ) + 73}{2 ( - 2 )} : - \frac{9 + 73}{4}

\frac{- ( - 9 ) + 73}{2 ( - 2 )}

Remove parentheses:

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

= \frac{9 + 73}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{9 + 73}{- 4}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{9 + 73}{4}

u = \frac{- ( - 9 ) - 73}{2 ( - 2 )} : - \frac{9 - 73}{4}

\frac{- ( - 9 ) - 73}{2 ( - 2 )}

Remove parentheses:

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

= \frac{9 - 73}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{9 - 73}{- 4}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{9 - 73}{4}

The solutions to the quadratic equation are:

u = - \frac{9 + 73}{4}, u = - \frac{9 - 73}{4}

Substitute back

u = sin (x)

sin (x) = - \frac{9 + 73}{4}, sin (x) = - \frac{9 - 73}{4}

sin (x) = - \frac{9 + 73}{4}, sin (x) = - \frac{9 - 73}{4}

sin (x) = - \frac{9 + 73}{4} :

No Solution

sin (x) = - \frac{9 + 73}{4}

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = - \frac{9 - 73}{4} : x = arcsin (- \frac{9 - 73}{4}) + 2 πn, x = π + arcsin (\frac{9 - 73}{4}) + 2 πn

sin (x) = - \frac{9 - 73}{4}

Apply trig inverse properties

sin (x) = - \frac{9 - 73}{4}

General solutions for

sin (x) = - \frac{9 - 73}{4}

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{9 - 73}{4}) + 2 πn, x = π + arcsin (\frac{9 - 73}{4}) + 2 πn

x = arcsin (- \frac{9 - 73}{4}) + 2 πn, x = π + arcsin (\frac{9 - 73}{4}) + 2 πn

Combine all the solutions

x = arcsin (- \frac{9 - 73}{4}) + 2 πn, x = π + arcsin (\frac{9 - 73}{4}) + 2 πn

Show solutions in decimal form

x = - 0.11424 \dots + 2 πn, x = π + 0.11424 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(x)-9sin(x)-3=0 ?
The general solution for 2cos^2(x)-9sin(x)-3=0 is x=-0.11424…+2pin,x=pi+0.11424…+2pin