What is the general solution for (3+sin(x))^2=13-cos^2(x) ?

The general solution for (3+sin(x))^2=13-cos^2(x) is x= pi/6+2pin,x=(5pi)/6+2pin

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(3+sin(x))^2=13-cos^2(x)

(3 + sin (x))^{2} = 13 - cos^{2} (x)

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

Degrees

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

Solution steps

(3 + sin (x))^{2} = 13 - cos^{2} (x)

Subtract

13 - cos^{2} (x)

from both sides

sin^{2} (x) + cos^{2} (x) + 6 sin (x) - 4 = 0

Rewrite using trig identities

sin^{2} (x) + cos^{2} (x) + 6 sin (x) - 4

Use the Pythagorean identity:

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

= - 4 + 6 sin (x) + 1

- 4 + 6 sin (x) + 1 = 0

Group like terms

6 sin (x) - 4 + 1 = 0

Add/Subtract the numbers:

- 4 + 1 = - 3

6 sin (x) - 3 = 0

Move

3

to the right side

6 sin (x) - 3 = 0

Add

3

to both sides

6 sin (x) - 3 + 3 = 0 + 3

Simplify

6 sin (x) = 3

6 sin (x) = 3

Divide both sides by

6

6 sin (x) = 3

Divide both sides by

6

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

Simplify

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

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

sec (x) = 3, \frac{3 π}{2} \leq x \leq 2 π

tan (2 x) = cot (3 x)

sin (x) = \infty

- e^{- s i n (x)} (sin (x) + cos^{2} (x)) = 0

cos^{2} (β) - cot^{2} (β) = cos^{2} (β) cot^{2} (β)

What is the general solution for (3+sin(x))^2=13-cos^2(x) ?
The general solution for (3+sin(x))^2=13-cos^2(x) is x= pi/6+2pin,x=(5pi)/6+2pin