What is the general solution for 4sin^2(x)+4sqrt(2)cos(x)-6=0 ?

The general solution for 4sin^2(x)+4sqrt(2)cos(x)-6=0 is x= pi/4+2pin,x=(7pi)/4+2pin

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4sin^2(x)+4sqrt(2)cos(x)-6=0

Solution

4 sin^{2} (x) + 42 cos (x) - 6 = 0

Solution

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Degrees

x = 4 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n

Solution steps

4 sin^{2} (x) + 42 cos (x) - 6 = 0

Rewrite using trig identities

- 6 + 4 sin^{2} (x) + 4 cos (x) 2

Use the Pythagorean identity:

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

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

= - 6 + 4 (1 - cos^{2} (x)) + 4 cos (x) 2

Simplify

- 6 + 4 (1 - cos^{2} (x)) + 4 cos (x) 2 : 42 cos (x) - 4 cos^{2} (x) - 2

- 6 + 4 (1 - cos^{2} (x)) + 4 cos (x) 2

= - 6 + 4 (1 - cos^{2} (x)) + 42 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)

= - 6 + 4 - 4 cos^{2} (x) + 4 cos (x) 2

Add/Subtract the numbers:

- 6 + 4 = - 2

= 42 cos (x) - 4 cos^{2} (x) - 2

= 42 cos (x) - 4 cos^{2} (x) - 2

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Write in the standard form

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

- 4 u^{2} + 42 u - 2 = 0

Solve with the quadratic formula

- 4 u^{2} + 42 u - 2 = 0

Quadratic Equation Formula:

For

a = - 4, b = 42, c = - 2

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

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

(42)^{2} - 4 (- 4) (- 2) = 0

(42)^{2} - 4 (- 4) (- 2)

Apply rule

- (- a) = a

= (42)^{2} - 4 \cdot 4 \cdot 2

(42)^{2} = 4^{2} \cdot 2

(42)^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

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

(2)^{2} : 2

Apply radical rule:

a = a^{\frac{1}{2}}

= (2^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 2^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

Cancel the common factor:

2

= 1

= 2

= 4^{2} \cdot 2

4 \cdot 4 \cdot 2 = 32

4 \cdot 4 \cdot 2

Multiply the numbers:

4 \cdot 4 \cdot 2 = 32

= 32

= 4^{2} \cdot 2 - 32

4^{2} \cdot 2 = 32

4^{2} \cdot 2

4^{2} = 16

= 16 \cdot 2

Multiply the numbers:

16 \cdot 2 = 32

= 32

= 32 - 32

Subtract the numbers:

32 - 32 = 0

= 0

u_{1, 2} = \frac{- 4 2 \pm 0}{2 ( - 4 )}

u = \frac{- 4 2}{2 ( - 4 )}

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

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

Remove parentheses:

(- a) = - a

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

Multiply the numbers:

2 \cdot 4 = 8

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

Apply the fraction rule:

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

= \frac{4 2}{8}

Cancel the common factor:

4

= \frac{2}{2}

u = \frac{2}{2}

The solution to the quadratic equation is:

u = \frac{2}{2}

Substitute back

u = cos (x)

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

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

cos (x) = \frac{2}{2} : x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

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

General solutions for

cos (x) = \frac{2}{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}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Combine all the solutions

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

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(tan(θ)cos(θ))/(csc(θ))=sin(θ)

\frac{tan ( θ ) cos ( θ )}{csc ( θ )} = sin (θ)

sin(θ)=(sqrt(44))/(12)

sin (θ) = \frac{44}{12}

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cos (θ) = - \frac{3}{5}, θ, 3

Frequently Asked Questions (FAQ)

What is the general solution for 4sin^2(x)+4sqrt(2)cos(x)-6=0 ?
The general solution for 4sin^2(x)+4sqrt(2)cos(x)-6=0 is x= pi/4+2pin,x=(7pi)/4+2pin