What is the general solution for 3sin^2(x)-4cos(x)+1=0 ?

The general solution for 3sin^2(x)-4cos(x)+1=0 is x=0.84106…+2pin,x=2pi-0.84106…+2pin

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3sin^2(x)-4cos(x)+1=0

Solution

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

Solution

x = 0.84106 \dots + 2 πn, x = 2 π - 0.84106 \dots + 2 πn

Degrees

x = 48.18968 \dots^{\circ} + 36 0^{\circ} n, x = 311.81031 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Expand

3 (1 - cos^{2} (x)) : 3 - 3 cos^{2} (x)

3 (1 - cos^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 3, b = 1, c = cos^{2} (x)

= 3 \cdot 1 - 3 cos^{2} (x)

Multiply the numbers:

3 \cdot 1 = 3

= 3 - 3 cos^{2} (x)

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

Add the numbers:

1 + 3 = 4

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

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

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

Solve by substitution

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

Let:

cos (x) = u

4 - 3 u^{2} - 4 u = 0

4 - 3 u^{2} - 4 u = 0 : u = - 2, u = \frac{2}{3}

4 - 3 u^{2} - 4 u = 0

Write in the standard form

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

- 3 u^{2} - 4 u + 4 = 0

Solve with the quadratic formula

- 3 u^{2} - 4 u + 4 = 0

Quadratic Equation Formula:

For

a = - 3, b = - 4, c = 4

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

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

(- 4)^{2} - 4 (- 3) \cdot 4 = 8

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

= 4^{2} + 4 \cdot 3 \cdot 4

Multiply the numbers:

4 \cdot 3 \cdot 4 = 48

= 4^{2} + 48

4^{2} = 16

= 16 + 48

Add the numbers:

16 + 48 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

u_{1, 2} = \frac{- ( - 4 ) \pm 8}{2 ( - 3 )}

Separate the solutions

u_{1} = \frac{- ( - 4 ) + 8}{2 ( - 3 )}, u_{2} = \frac{- ( - 4 ) - 8}{2 ( - 3 )}

u = \frac{- ( - 4 ) + 8}{2 ( - 3 )} : - 2

\frac{- ( - 4 ) + 8}{2 ( - 3 )}

Remove parentheses:

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

= \frac{4 + 8}{- 2 \cdot 3}

Add the numbers:

4 + 8 = 12

= \frac{12}{- 2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{12}{- 6}

Apply the fraction rule:

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

= - \frac{12}{6}

Divide the numbers:

\frac{12}{6} = 2

= - 2

u = \frac{- ( - 4 ) - 8}{2 ( - 3 )} : \frac{2}{3}

\frac{- ( - 4 ) - 8}{2 ( - 3 )}

Remove parentheses:

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

= \frac{4 - 8}{- 2 \cdot 3}

Subtract the numbers:

4 - 8 = - 4

= \frac{- 4}{- 2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 4}{- 6}

Apply the fraction rule:

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

= \frac{4}{6}

Cancel the common factor:

2

= \frac{2}{3}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

cos (x) = - 2, cos (x) = \frac{2}{3}

cos (x) = - 2, cos (x) = \frac{2}{3}

cos (x) = - 2 :

No Solution

cos (x) = - 2

- 1 \leq cos (x) \leq 1

No Solution

cos (x) = \frac{2}{3} : x = arccos (\frac{2}{3}) + 2 πn, x = 2 π - arccos (\frac{2}{3}) + 2 πn

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

Apply trig inverse properties

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

General solutions for

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

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{2}{3}) + 2 πn, x = 2 π - arccos (\frac{2}{3}) + 2 πn

x = arccos (\frac{2}{3}) + 2 πn, x = 2 π - arccos (\frac{2}{3}) + 2 πn

Combine all the solutions

x = arccos (\frac{2}{3}) + 2 πn, x = 2 π - arccos (\frac{2}{3}) + 2 πn

Show solutions in decimal form

x = 0.84106 \dots + 2 πn, x = 2 π - 0.84106 \dots + 2 πn

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sin(-pi/8 c)=-14/14

sin (- \frac{π}{8} c) = - \frac{14}{14}

(25)/(sin(x))=(18)/(sin(36))

\frac{25}{sin ( x )} = \frac{18}{sin ( 3 6 ^{\circ} )}

Frequently Asked Questions (FAQ)

What is the general solution for 3sin^2(x)-4cos(x)+1=0 ?
The general solution for 3sin^2(x)-4cos(x)+1=0 is x=0.84106…+2pin,x=2pi-0.84106…+2pin