What is the general solution for (-6cos(x)+2sin(x))^2+32sin^2(x)=48 ?

The general solution for (-6cos(x)+2sin(x))^2+32sin^2(x)=48 is x=(3pi)/4+pin

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(-6cos(x)+2sin(x))^2+32sin^2(x)=48

(- 6 cos (x) + 2 sin (x))^{2} + 32 sin^{2} (x) = 48

x = \frac{3 π}{4} + πn

Solution steps

(- 6 cos (x) + 2 sin (x))^{2} + 32 sin^{2} (x) = 48

Subtract

48

from both sides

36 cos^{2} (x) - 24 cos (x) sin (x) + 36 sin^{2} (x) - 48 = 0

Rewrite using trig identities

- 12 - 24 \cdot \frac{sin ( 2 x )}{2} = 0

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

- 12 - 12 sin (2 x) = 0

Move

12

to the right side

- 12 sin (2 x) = 12

Divide both sides by

- 12

sin (2 x) = - 1

General solutions for

sin (2 x) = - 1

2 x = \frac{3 π}{2} + 2 πn

Solve

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

x = \frac{3 π}{4} + πn

tan (x) + 5 = \frac{15 - 3}{3}

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

tan (x) csc (x) - 2 tan (x) = 3 csc (x) - 6

x, sin (x) = ln (y)

10 = 7 cos ((\frac{π}{6}) (t + 6)) + 8

What is the general solution for (-6cos(x)+2sin(x))^2+32sin^2(x)=48 ?
The general solution for (-6cos(x)+2sin(x))^2+32sin^2(x)=48 is x=(3pi)/4+pin