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

The general solution for cos^2(x)=(3(1-sin(x)))/2 is x= pi/6+2pin,x=(5pi)/6+2pin,x= pi/2+2pin

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cos^2(x)=(3(1-sin(x)))/2

Solution

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

Solution

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{π}{2} + 2 πn

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

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

from both sides

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

- (1 - sin (x)) \cdot 3 + 2 (1 - sin^{2} (x)) : 3 sin (x) - 2 sin^{2} (x) - 1

- (1 - sin (x)) \cdot 3 + 2 (1 - sin^{2} (x))

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

Expand

- 3 (1 - sin (x)) : - 3 + 3 sin (x)

- 3 (1 - sin (x))

Apply the distributive law:

a (b - c) = ab - a c

a = - 3, b = 1, c = sin (x)

= - 3 \cdot 1 - (- 3) sin (x)

Apply minus-plus rules

- (- a) = a

= - 3 \cdot 1 + 3 sin (x)

Multiply the numbers:

3 \cdot 1 = 3

= - 3 + 3 sin (x)

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

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

Simplify

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

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

Group like terms

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

Add/Subtract the numbers:

- 3 + 2 = - 1

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

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

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

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

Solve by substitution

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

Let:

sin (x) = u

- 1 - 2 u^{2} + 3 u = 0

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

- 1 - 2 u^{2} + 3 u = 0

Write in the standard form

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

- 2 u^{2} + 3 u - 1 = 0

Solve with the quadratic formula

- 2 u^{2} + 3 u - 1 = 0

Quadratic Equation Formula:

For

a = - 2, b = 3, c = - 1

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

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

3^{2} - 4 (- 2) (- 1) = 1

3^{2} - 4 (- 2) (- 1)

Apply rule

- (- a) = a

= 3^{2} - 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

u = \frac{- 3 + 1}{2 ( - 2 )} : \frac{1}{2}

\frac{- 3 + 1}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 3 + 1}{- 2 \cdot 2}

Add/Subtract the numbers:

- 3 + 1 = - 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

Apply the fraction rule:

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

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

u = \frac{- 3 - 1}{2 ( - 2 )} : 1

\frac{- 3 - 1}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 3 - 1}{- 2 \cdot 2}

Subtract the numbers:

- 3 - 1 = - 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{- 4}

Apply the fraction rule:

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

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

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

Substitute back

u = sin (x)

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

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

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

General solutions for

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

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

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

General solutions for

sin (x) = 1

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

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{π}{2} + 2 πn

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9.8 \cdot sin (α) - 1.96 \cdot cos (α) = 0.47

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2-5cos(x+1)=4

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Frequently Asked Questions (FAQ)

What is the general solution for cos^2(x)=(3(1-sin(x)))/2 ?
The general solution for cos^2(x)=(3(1-sin(x)))/2 is x= pi/6+2pin,x=(5pi)/6+2pin,x= pi/2+2pin