What is the general solution for sin(3x)+1=cos(3x) ?

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

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sin(3x)+1=cos(3x)

sin (3 x) + 1 = cos (3 x)

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

Solution steps

sin (3 x) + 1 = cos (3 x)

Subtract

cos (3 x)

from both sides

sin (3 x) + 1 - cos (3 x) = 0

Rewrite using trig identities

1 + 2 sin (3 x - \frac{π}{4}) = 0

Move

1

to the right side

2 sin (3 x - \frac{π}{4}) = - 1

Divide both sides by

2

sin (3 x - \frac{π}{4}) = - \frac{2}{2}

General solutions for

sin (3 x - \frac{π}{4}) = - \frac{2}{2}

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

Solve

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

Solve

3 x - \frac{π}{4} = \frac{7 π}{4} + 2 πn : x = \frac{2 πn}{3} + \frac{2 π}{3}

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

6 sin^{2} (x) = 5

tan (θ) = \frac{46}{33}

arcsin (y) = \frac{1}{2}

500 = 2 cos (x)

cot (3 x) = 1, 0 \leq x \leq 2 π

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