What is the general solution for 4sin^2(x)+8cos(x)-7=0 ?

The general solution for 4sin^2(x)+8cos(x)-7=0 is x= pi/3+2pin,x=(5pi)/3+2pin

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4sin^2(x)+8cos(x)-7=0

Solution

4 sin^{2} (x) + 8 cos (x) - 7 = 0

Solution

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Degrees

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

4 sin^{2} (x) + 8 cos (x) - 7 = 0

Rewrite using trig identities

- 7 + 4 sin^{2} (x) + 8 cos (x)

Use the Pythagorean identity:

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

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

= - 7 + 4 (1 - cos^{2} (x)) + 8 cos (x)

Simplify

- 7 + 4 (1 - cos^{2} (x)) + 8 cos (x) : 8 cos (x) - 4 cos^{2} (x) - 3

- 7 + 4 (1 - cos^{2} (x)) + 8 cos (x)

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Multiply the numbers:

4 \cdot 1 = 4

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

= - 7 + 4 - 4 cos^{2} (x) + 8 cos (x)

Add/Subtract the numbers:

- 7 + 4 = - 3

= 8 cos (x) - 4 cos^{2} (x) - 3

= 8 cos (x) - 4 cos^{2} (x) - 3

- 3 - 4 cos^{2} (x) + 8 cos (x) = 0

Solve by substitution

- 3 - 4 cos^{2} (x) + 8 cos (x) = 0

Let:

cos (x) = u

- 3 - 4 u^{2} + 8 u = 0

- 3 - 4 u^{2} + 8 u = 0 : u = \frac{1}{2}, u = \frac{3}{2}

- 3 - 4 u^{2} + 8 u = 0

Write in the standard form

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

- 4 u^{2} + 8 u - 3 = 0

Solve with the quadratic formula

- 4 u^{2} + 8 u - 3 = 0

Quadratic Equation Formula:

For

a = - 4, b = 8, c = - 3

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

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

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

8^{2} - 4 (- 4) (- 3)

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 4 \cdot 3 = 48

= 8^{2} - 48

8^{2} = 64

= 64 - 48

Subtract the numbers:

64 - 48 = 16

= 16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

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

Separate the solutions

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

u = \frac{- 8 + 4}{2 ( - 4 )} : \frac{1}{2}

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

Remove parentheses:

(- a) = - a

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

Add/Subtract the numbers:

- 8 + 4 = - 4

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 4}{- 8}

Apply the fraction rule:

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

= \frac{4}{8}

Cancel the common factor:

4

= \frac{1}{2}

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

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

Remove parentheses:

(- a) = - a

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

Subtract the numbers:

- 8 - 4 = - 12

= \frac{- 12}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 12}{- 8}

Apply the fraction rule:

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

= \frac{12}{8}

Cancel the common factor:

4

= \frac{3}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = \frac{1}{2}

General solutions for

cos (x) = \frac{1}{2}

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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

No Solution

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

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Popular Examples

6(cos(b))^2+cos(b)-1=0

6 (cos (b))^{2} + cos (b) - 1 = 0

sin(θ)=(1.33)/(2.42)

sin (θ) = \frac{1.33}{2.42}

2tan(x)=3-tan^2(x),-2pi<= ,x<= pi

2 tan (x) = 3 - tan^{2} (x), - 2 π \leq, x \leq π

-25=(29cos(x))*3.7-4.9*3.7^2

- 25 = (29 cos (x)) \cdot 3.7 - 4.9 \cdot 3. 7^{2}

solvefor t,x=5(sin(t)+cos(t))solve for

t, x = 5 (sin (t) + cos (t))

Frequently Asked Questions (FAQ)

What is the general solution for 4sin^2(x)+8cos(x)-7=0 ?
The general solution for 4sin^2(x)+8cos(x)-7=0 is x= pi/3+2pin,x=(5pi)/3+2pin