What is the general solution for 4cos(2x)+2sin(x)=1,0<= x<= 360 ?

The general solution for 4cos(2x)+2sin(x)=1,0<= x<= 360 is x=210,x=330,x=0.84806…,x=180-0.84806…

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4cos(2x)+2sin(x)=1,0<= x<= 360

Solution

4 cos (2 x) + 2 sin (x) = 1, 0^{\circ} \leq x \leq 36 0^{\circ}

Solution

x = 21 0^{\circ}, x = 33 0^{\circ}, x = 0.84806 \dots, x = 18 0^{\circ} - 0.84806 \dots

Radians

x = \frac{7 π}{6}, x = \frac{11 π}{6}, x = 0.84806 \dots, x = π - 0.84806 \dots

Solution steps

4 cos (2 x) + 2 sin (x) = 1, 0^{\circ} \leq x \leq 36 0^{\circ}

Subtract

1

from both sides

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

Rewrite using trig identities

- 1 + 2 sin (x) + 4 cos (2 x)

Use the Double Angle identity:

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

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

Simplify

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

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

Expand

4 (1 - 2 sin^{2} (x)) : 4 - 8 sin^{2} (x)

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Simplify

4 \cdot 1 - 4 \cdot 2 sin^{2} (x) : 4 - 8 sin^{2} (x)

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

Multiply the numbers:

4 \cdot 1 = 4

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

Multiply the numbers:

4 \cdot 2 = 8

= 4 - 8 sin^{2} (x)

= 4 - 8 sin^{2} (x)

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

Simplify

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

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

Group like terms

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

Add/Subtract the numbers:

- 1 + 4 = 3

= 2 sin (x) - 8 sin^{2} (x) + 3

= 2 sin (x) - 8 sin^{2} (x) + 3

= 2 sin (x) - 8 sin^{2} (x) + 3

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 8, b = 2, c = 3

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

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

2^{2} - 4 (- 8) \cdot 3 = 10

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 8 \cdot 3 = 96

= 2^{2} + 96

2^{2} = 4

= 4 + 96

Add the numbers:

4 + 96 = 100

= 100

Factor the number:

100 = 1 0^{2}

= 1 0^{2}

Apply radical rule:

1 0^{2} = 10

= 10

u_{1, 2} = \frac{- 2 \pm 10}{2 ( - 8 )}

Separate the solutions

u_{1} = \frac{- 2 + 10}{2 ( - 8 )}, u_{2} = \frac{- 2 - 10}{2 ( - 8 )}

u = \frac{- 2 + 10}{2 ( - 8 )} : - \frac{1}{2}

\frac{- 2 + 10}{2 ( - 8 )}

Remove parentheses:

(- a) = - a

= \frac{- 2 + 10}{- 2 \cdot 8}

Add/Subtract the numbers:

- 2 + 10 = 8

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

Multiply the numbers:

2 \cdot 8 = 16

= \frac{8}{- 16}

Apply the fraction rule:

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

= - \frac{8}{16}

Cancel the common factor:

8

= - \frac{1}{2}

u = \frac{- 2 - 10}{2 ( - 8 )} : \frac{3}{4}

\frac{- 2 - 10}{2 ( - 8 )}

Remove parentheses:

(- a) = - a

= \frac{- 2 - 10}{- 2 \cdot 8}

Subtract the numbers:

- 2 - 10 = - 12

= \frac{- 12}{- 2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 12}{- 16}

Apply the fraction rule:

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

= \frac{12}{16}

Cancel the common factor:

4

= \frac{3}{4}

The solutions to the quadratic equation are:

u = - \frac{1}{2}, u = \frac{3}{4}

Substitute back

u = sin (x)

sin (x) = - \frac{1}{2}, sin (x) = \frac{3}{4}

sin (x) = - \frac{1}{2}, sin (x) = \frac{3}{4}

sin (x) = - \frac{1}{2}, 0 \leq x \leq 36 0^{\circ} : x = 21 0^{\circ}, x = 33 0^{\circ}

sin (x) = - \frac{1}{2}, 0 \leq x \leq 36 0^{\circ}

General solutions for

sin (x) = - \frac{1}{2}

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 21 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n

x = 21 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n

Solutions for the range

0 \leq x \leq 36 0^{\circ}

x = 21 0^{\circ}, x = 33 0^{\circ}

sin (x) = \frac{3}{4}, 0 \leq x \leq 36 0^{\circ} : x = arcsin (\frac{3}{4}), x = 18 0^{\circ} - arcsin (\frac{3}{4})

sin (x) = \frac{3}{4}, 0 \leq x \leq 36 0^{\circ}

Apply trig inverse properties

sin (x) = \frac{3}{4}

General solutions for

sin (x) = \frac{3}{4}

sin (x) = a \Rightarrow x = arcsin (a) + 36 0^{\circ} n, x = 18 0^{\circ} - arcsin (a) + 36 0^{\circ} n

x = arcsin (\frac{3}{4}) + 36 0^{\circ} n, x = 18 0^{\circ} - arcsin (\frac{3}{4}) + 36 0^{\circ} n

x = arcsin (\frac{3}{4}) + 36 0^{\circ} n, x = 18 0^{\circ} - arcsin (\frac{3}{4}) + 36 0^{\circ} n

Solutions for the range

0 \leq x \leq 36 0^{\circ}

x = arcsin (\frac{3}{4}), x = 18 0^{\circ} - arcsin (\frac{3}{4})

Combine all the solutions

x = 21 0^{\circ}, x = 33 0^{\circ}, x = arcsin (\frac{3}{4}), x = 18 0^{\circ} - arcsin (\frac{3}{4})

Show solutions in decimal form

x = 21 0^{\circ}, x = 33 0^{\circ}, x = 0.84806 \dots, x = 18 0^{\circ} - 0.84806 \dots

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Frequently Asked Questions (FAQ)

What is the general solution for 4cos(2x)+2sin(x)=1,0<= x<= 360 ?
The general solution for 4cos(2x)+2sin(x)=1,0<= x<= 360 is x=210,x=330,x=0.84806…,x=180-0.84806…