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Popular >

3-2cos^2(x)=4sin^2(x)

Solution

3 - 2 cos^{2} (x) = 4 sin^{2} (x)

Solution

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Degrees

x = 4 5^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n, x = 22 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n

Solution steps

3 - 2 cos^{2} (x) = 4 sin^{2} (x)

Subtract

4 sin^{2} (x)

from both sides

3 - 2 cos^{2} (x) - 4 sin^{2} (x) = 0

Rewrite using trig identities

3 - 2 cos^{2} (x) - 4 sin^{2} (x)

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Apply minus-plus rules

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

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

= 3 - 2 + 2 sin^{2} (x) - 4 sin^{2} (x)

Simplify

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

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

Add similar elements:

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

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

Subtract the numbers:

3 - 2 = 1

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

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

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

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

Solve by substitution

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

Let:

sin (x) = u

1 - 2 u^{2} = 0

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

1 - 2 u^{2} = 0

Move

1

to the right side

1 - 2 u^{2} = 0

Subtract

1

from both sides

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

Simplify

- 2 u^{2} = - 1

- 2 u^{2} = - 1

Divide both sides by

- 2

- 2 u^{2} = - 1

Divide both sides by

- 2

\frac{- 2 u ^{2}}{- 2} = \frac{- 1}{- 2}

Simplify

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

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

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

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

Substitute back

u = sin (x)

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

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

sin (x) = \frac{1}{2} : x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 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}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

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

sin (x) = - \frac{1}{2} : x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 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{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Combine all the solutions

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

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