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

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

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3cos(2x)+2sin^2(x)=2

Solution

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

Solution

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

Degrees

x = 3 0^{\circ} + 18 0^{\circ} n, x = 15 0^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

2

from both sides

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

= - 2 - (cos (2 x) - 1) + 3 cos (2 x)

Simplify

- 2 - (cos (2 x) - 1) + 3 cos (2 x) : 2 cos (2 x) - 1

- 2 - (cos (2 x) - 1) + 3 cos (2 x)

- (cos (2 x) - 1) : - cos (2 x) + 1

- (cos (2 x) - 1)

Distribute parentheses

= - (cos (2 x)) - (- 1)

Apply minus-plus rules

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

= - cos (2 x) + 1

= - 2 - cos (2 x) + 1 + 3 cos (2 x)

Simplify

- 2 - cos (2 x) + 1 + 3 cos (2 x) : 2 cos (2 x) - 1

- 2 - cos (2 x) + 1 + 3 cos (2 x)

Group like terms

= - cos (2 x) + 3 cos (2 x) - 2 + 1

Add similar elements:

- cos (2 x) + 3 cos (2 x) = 2 cos (2 x)

= 2 cos (2 x) - 2 + 1

Add/Subtract the numbers:

- 2 + 1 = - 1

= 2 cos (2 x) - 1

= 2 cos (2 x) - 1

= 2 cos (2 x) - 1

- 1 + 2 cos (2 x) = 0

Move

1

to the right side

- 1 + 2 cos (2 x) = 0

Add

1

to both sides

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

Simplify

2 cos (2 x) = 1

2 cos (2 x) = 1

Divide both sides by

2

2 cos (2 x) = 1

Divide both sides by

2

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

Simplify

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

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

General solutions for

cos (2 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}

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

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

Solve

2 x = \frac{π}{3} + 2 πn : x = \frac{π}{6} + πn

2 x = \frac{π}{3} + 2 πn

Divide both sides by

2

2 x = \frac{π}{3} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{π}{3}}{2} + \frac{2 πn}{2} : \frac{π}{6} + πn

\frac{\frac{π}{3}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{3}}{2} = \frac{π}{6}

\frac{\frac{π}{3}}{2}

Apply the fraction rule:

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

= \frac{π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{π}{6}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{6} + πn

x = \frac{π}{6} + πn

x = \frac{π}{6} + πn

x = \frac{π}{6} + πn

Solve

2 x = \frac{5 π}{3} + 2 πn : x = \frac{5 π}{6} + πn

2 x = \frac{5 π}{3} + 2 πn

Divide both sides by

2

2 x = \frac{5 π}{3} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{5 π}{3}}{2} + \frac{2 πn}{2} : \frac{5 π}{6} + πn

\frac{\frac{5 π}{3}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{3}}{2} = \frac{5 π}{6}

\frac{\frac{5 π}{3}}{2}

Apply the fraction rule:

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

= \frac{5 π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{5 π}{6}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{5 π}{6} + πn

x = \frac{5 π}{6} + πn

x = \frac{5 π}{6} + πn

x = \frac{5 π}{6} + πn

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

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Frequently Asked Questions (FAQ)

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