What is the general solution for 1+sin(2x)=5(sin(x)+cos(x)) ?

The general solution for 1+sin(2x)=5(sin(x)+cos(x)) is x=(3pi)/4+pin

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1+sin(2x)=5(sin(x)+cos(x))

1 + sin (2 x) = 5 (sin (x) + cos (x))

x = \frac{3 π}{4} + πn

Solution steps

1 + sin (2 x) = 5 (sin (x) + cos (x))

Subtract

5 (sin (x) + cos (x))

from both sides

1 + sin (2 x) - 5 (sin (x) + cos (x)) = 0

Rewrite using trig identities

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

Factor

(cos (x) + sin (x))^{2} - (cos (x) + sin (x)) \cdot 5 : (cos (x) + sin (x)) (cos (x) + sin (x) - 5)

(cos (x) + sin (x)) (cos (x) + sin (x) - 5) = 0

Solving each part separately

cos (x) + sin (x) = 0 or cos (x) + sin (x) - 5 = 0

cos (x) + sin (x) = 0 : x = \frac{3 π}{4} + πn

cos (x) + sin (x) - 5 = 0 :

No Solution

Combine all the solutions

x = \frac{3 π}{4} + πn

x, y = I n [tan (5 x^{2} + 3 x + 2)]

cos (2 x) = \frac{3}{4} - 5 cos^{2} (x)

sin (x + \frac{π}{9}) = \frac{3}{2}

3 cos (2 θ) = 3 sin (θ)

cos^{2} (x) = 1 - cos^{2} (x), [36 0^{\circ}]

What is the general solution for 1+sin(2x)=5(sin(x)+cos(x)) ?
The general solution for 1+sin(2x)=5(sin(x)+cos(x)) is x=(3pi)/4+pin