What is the general solution for (3cos(x)+2sin(x))^2+5sin^2(x)=9-3sqrt(3) ?

The general solution for (3cos(x)+2sin(x))^2+5sin^2(x)=9-3sqrt(3) is x=(2pi)/3+pin,x=(5pi)/6+pin

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(3cos(x)+2sin(x))^2+5sin^2(x)=9-3sqrt(3)

Solution

(3 cos (x) + 2 sin (x))^{2} + 5 sin^{2} (x) = 9 - 33

Solution

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

Degrees

x = 12 0^{\circ} + 18 0^{\circ} n, x = 15 0^{\circ} + 18 0^{\circ} n

Solution steps

(3 cos (x) + 2 sin (x))^{2} + 5 sin^{2} (x) = 9 - 33

Subtract

9 - 33

from both sides

9 cos^{2} (x) + 12 cos (x) sin (x) + 9 sin^{2} (x) - 9 + 33 = 0

Rewrite using trig identities

- 9 + 33 + 9 cos^{2} (x) + 9 sin^{2} (x) + 12 cos (x) sin (x)

Use the Pythagorean identity:

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

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

= - 9 + 33 + 9 (1 - sin^{2} (x)) + 9 sin^{2} (x) + 12 cos (x) sin (x)

Simplify

- 9 + 33 + 9 (1 - sin^{2} (x)) + 9 sin^{2} (x) + 12 cos (x) sin (x) : 12 cos (x) sin (x) + 33

- 9 + 33 + 9 (1 - sin^{2} (x)) + 9 sin^{2} (x) + 12 cos (x) sin (x)

Expand

9 (1 - sin^{2} (x)) : 9 - 9 sin^{2} (x)

9 (1 - sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 9, b = 1, c = sin^{2} (x)

= 9 \cdot 1 - 9 sin^{2} (x)

Multiply the numbers:

9 \cdot 1 = 9

= 9 - 9 sin^{2} (x)

= - 9 + 33 + 9 - 9 sin^{2} (x) + 9 sin^{2} (x) + 12 cos (x) sin (x)

Simplify

- 9 + 33 + 9 - 9 sin^{2} (x) + 9 sin^{2} (x) + 12 cos (x) sin (x) : 12 cos (x) sin (x) + 33

- 9 + 33 + 9 - 9 sin^{2} (x) + 9 sin^{2} (x) + 12 cos (x) sin (x)

Add similar elements:

- 9 sin^{2} (x) + 9 sin^{2} (x) = 0

= - 9 + 33 + 9 + 12 cos (x) sin (x)

Group like terms

= 12 cos (x) sin (x) - 9 + 33 + 9

- 9 + 9 = 0

= 12 cos (x) sin (x) + 33

= 12 cos (x) sin (x) + 33

= 12 cos (x) sin (x) + 33

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

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

= 33 + 12 \cdot \frac{sin ( 2 x )}{2}

33 + 12 \cdot \frac{sin ( 2 x )}{2} = 0

12 \cdot \frac{sin ( 2 x )}{2} = 6 sin (2 x)

12 \cdot \frac{sin ( 2 x )}{2}

Multiply fractions:

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

= \frac{sin ( 2 x ) \cdot 12}{2}

Divide the numbers:

\frac{12}{2} = 6

= 6 sin (2 x)

33 + 6 sin (2 x) = 0

Move

33

to the right side

33 + 6 sin (2 x) = 0

Subtract

33

from both sides

33 + 6 sin (2 x) - 33 = 0 - 33

Simplify

6 sin (2 x) = - 33

6 sin (2 x) = - 33

Divide both sides by

6

6 sin (2 x) = - 33

Divide both sides by

6

\frac{6 sin ( 2 x )}{6} = \frac{- 3 3}{6}

Simplify

\frac{6 sin ( 2 x )}{6} = \frac{- 3 3}{6}

Simplify

\frac{6 sin ( 2 x )}{6} : sin (2 x)

\frac{6 sin ( 2 x )}{6}

Divide the numbers:

\frac{6}{6} = 1

= sin (2 x)

Simplify

\frac{- 3 3}{6} : - \frac{3}{2}

\frac{- 3 3}{6}

Apply the fraction rule:

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

= - \frac{3 3}{6}

Cancel the common factor:

3

= - \frac{3}{2}

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

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

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

General solutions for

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

2 x = \frac{4 π}{3} + 2 πn, 2 x = \frac{5 π}{3} + 2 πn

2 x = \frac{4 π}{3} + 2 πn, 2 x = \frac{5 π}{3} + 2 πn

Solve

2 x = \frac{4 π}{3} + 2 πn : x = \frac{2 π}{3} + πn

2 x = \frac{4 π}{3} + 2 πn

Divide both sides by

2

2 x = \frac{4 π}{3} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{4 π}{3}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{4 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{4 π}{3}}{2} + \frac{2 πn}{2} : \frac{2 π}{3} + πn

\frac{\frac{4 π}{3}}{2} + \frac{2 πn}{2}

\frac{\frac{4 π}{3}}{2} = \frac{2 π}{3}

\frac{\frac{4 π}{3}}{2}

Apply the fraction rule:

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

= \frac{4 π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{4 π}{6}

Cancel the common factor:

2

= \frac{2 π}{3}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{2 π}{3} + πn

x = \frac{2 π}{3} + πn

x = \frac{2 π}{3} + πn

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

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Frequently Asked Questions (FAQ)

What is the general solution for (3cos(x)+2sin(x))^2+5sin^2(x)=9-3sqrt(3) ?
The general solution for (3cos(x)+2sin(x))^2+5sin^2(x)=9-3sqrt(3) is x=(2pi)/3+pin,x=(5pi)/6+pin