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Popular >

8sin^2(x)+4cos^2(x)=7

Solution

8 sin^{2} (x) + 4 cos^{2} (x) = 7

Solution

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

Degrees

x = 6 0^{\circ} + 36 0^{\circ} n, x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

8 sin^{2} (x) + 4 cos^{2} (x) = 7

Subtract

7

from both sides

8 sin^{2} (x) + 4 cos^{2} (x) - 7 = 0

Rewrite using trig identities

- 7 + 4 cos^{2} (x) + 8 sin^{2} (x)

Use the Pythagorean identity:

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

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

= - 7 + 4 (1 - sin^{2} (x)) + 8 sin^{2} (x)

Simplify

- 7 + 4 (1 - sin^{2} (x)) + 8 sin^{2} (x) : 4 sin^{2} (x) - 3

- 7 + 4 (1 - sin^{2} (x)) + 8 sin^{2} (x)

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Multiply the numbers:

4 \cdot 1 = 4

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

= - 7 + 4 - 4 sin^{2} (x) + 8 sin^{2} (x)

Simplify

- 7 + 4 - 4 sin^{2} (x) + 8 sin^{2} (x) : 4 sin^{2} (x) - 3

- 7 + 4 - 4 sin^{2} (x) + 8 sin^{2} (x)

Add similar elements:

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

= - 7 + 4 + 4 sin^{2} (x)

Add/Subtract the numbers:

- 7 + 4 = - 3

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

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

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

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

Solve by substitution

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

Let:

sin (x) = u

- 3 + 4 u^{2} = 0

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

- 3 + 4 u^{2} = 0

Move

3

to the right side

- 3 + 4 u^{2} = 0

Add

3

to both sides

- 3 + 4 u^{2} + 3 = 0 + 3

Simplify

4 u^{2} = 3

4 u^{2} = 3

Divide both sides by

4

4 u^{2} = 3

Divide both sides by

4

\frac{4 u ^{2}}{4} = \frac{3}{4}

Simplify

u^{2} = \frac{3}{4}

u^{2} = \frac{3}{4}

For

x^{2} = f (a)

the solutions are

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

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

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

\frac{3}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{3}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{3}{2}

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

- \frac{3}{4}

Simplify

\frac{3}{4} : \frac{3}{2}

\frac{3}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{3}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{3}{2}

= - \frac{3}{2}

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

Substitute back

u = sin (x)

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

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

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

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

General solutions for

sin (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}

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

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

sin (x) = - \frac{3}{2} : x = \frac{4 π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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

General solutions for

sin (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}

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

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

Combine all the solutions

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

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