What is the general solution for 4cos^2(x)+5sin(x)=3 ?

The general solution for 4cos^2(x)+5sin(x)=3 is x=-0.17630…+2pin,x=pi+0.17630…+2pin

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Popular >

4cos^2(x)+5sin(x)=3

Solution

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

Solution

x = - 0.17630 \dots + 2 πn, x = π + 0.17630 \dots + 2 πn

Degrees

x = - 10.10138 \dots^{\circ} + 36 0^{\circ} n, x = 190.10138 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

3

from both sides

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

Rewrite using trig identities

- 3 + 4 cos^{2} (x) + 5 sin (x)

Use the Pythagorean identity:

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

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

= - 3 + 4 (1 - sin^{2} (x)) + 5 sin (x)

Simplify

- 3 + 4 (1 - sin^{2} (x)) + 5 sin (x) : 5 sin (x) - 4 sin^{2} (x) + 1

- 3 + 4 (1 - sin^{2} (x)) + 5 sin (x)

Expand

4 (1 - sin^{2} (x)) : 4 - 4 sin^{2} (x)

4 (1 - sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 1, c = sin^{2} (x)

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

Multiply the numbers:

4 \cdot 1 = 4

= 4 - 4 sin^{2} (x)

= - 3 + 4 - 4 sin^{2} (x) + 5 sin (x)

Add/Subtract the numbers:

- 3 + 4 = 1

= 5 sin (x) - 4 sin^{2} (x) + 1

= 5 sin (x) - 4 sin^{2} (x) + 1

1 - 4 sin^{2} (x) + 5 sin (x) = 0

Solve by substitution

1 - 4 sin^{2} (x) + 5 sin (x) = 0

Let:

sin (x) = u

1 - 4 u^{2} + 5 u = 0

1 - 4 u^{2} + 5 u = 0 : u = - \frac{- 5 + 41}{8}, u = \frac{5 + 41}{8}

1 - 4 u^{2} + 5 u = 0

Write in the standard form

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

- 4 u^{2} + 5 u + 1 = 0

Solve with the quadratic formula

- 4 u^{2} + 5 u + 1 = 0

Quadratic Equation Formula:

For

a = - 4, b = 5, c = 1

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

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

5^{2} - 4 (- 4) \cdot 1 = 41

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

Apply rule

- (- a) = a

= 5^{2} + 4 \cdot 4 \cdot 1

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 5^{2} + 16

5^{2} = 25

= 25 + 16

Add the numbers:

25 + 16 = 41

= 41

u_{1, 2} = \frac{- 5 \pm 41}{2 ( - 4 )}

Separate the solutions

u_{1} = \frac{- 5 + 41}{2 ( - 4 )}, u_{2} = \frac{- 5 - 41}{2 ( - 4 )}

u = \frac{- 5 + 41}{2 ( - 4 )} : - \frac{- 5 + 41}{8}

\frac{- 5 + 41}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

= \frac{- 5 + 41}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 5 + 41}{- 8}

Apply the fraction rule:

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

= - \frac{- 5 + 41}{8}

u = \frac{- 5 - 41}{2 ( - 4 )} : \frac{5 + 41}{8}

\frac{- 5 - 41}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

= \frac{- 5 - 41}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 5 - 41}{- 8}

Apply the fraction rule:

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

- 5 - 41 = - (5 + 41)

= \frac{5 + 41}{8}

The solutions to the quadratic equation are:

u = - \frac{- 5 + 41}{8}, u = \frac{5 + 41}{8}

Substitute back

u = sin (x)

sin (x) = - \frac{- 5 + 41}{8}, sin (x) = \frac{5 + 41}{8}

sin (x) = - \frac{- 5 + 41}{8}, sin (x) = \frac{5 + 41}{8}

sin (x) = - \frac{- 5 + 41}{8} : x = arcsin (- \frac{- 5 + 41}{8}) + 2 πn, x = π + arcsin (\frac{- 5 + 41}{8}) + 2 πn

sin (x) = - \frac{- 5 + 41}{8}

Apply trig inverse properties

sin (x) = - \frac{- 5 + 41}{8}

General solutions for

sin (x) = - \frac{- 5 + 41}{8}

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

x = arcsin (- \frac{- 5 + 41}{8}) + 2 πn, x = π + arcsin (\frac{- 5 + 41}{8}) + 2 πn

x = arcsin (- \frac{- 5 + 41}{8}) + 2 πn, x = π + arcsin (\frac{- 5 + 41}{8}) + 2 πn

sin (x) = \frac{5 + 41}{8} :

No Solution

sin (x) = \frac{5 + 41}{8}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = arcsin (- \frac{- 5 + 41}{8}) + 2 πn, x = π + arcsin (\frac{- 5 + 41}{8}) + 2 πn

Show solutions in decimal form

x = - 0.17630 \dots + 2 πn, x = π + 0.17630 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 4cos^2(x)+5sin(x)=3 ?
The general solution for 4cos^2(x)+5sin(x)=3 is x=-0.17630…+2pin,x=pi+0.17630…+2pin