Solve for x on the interval [0, 2pi]
Sinx = cosx + 1

π

Step-by-step explanation:

Solve for x on the interval [0, 2pi]

Given the equation

Sinx = cosx + 1

Square both sides of the equation

Sin²x = (cos x + 1)²

Sin²x = cos²x + 2cos x + 1

Since Sin²x = 1 - cos²x

1 - cos²x = cos²x + 2cos x + 1

Collect like terms

1-1-cos²x-cos²x-2cos x  = 0

-2cos²x-2cos x = 0

-2cos²x = 2cos x

-cosx = 1

cos x = -1

x = arccos -1

x = 180 degrees

Hence the value of x = π

you need a common difference between consecutive terms to be constant, like 1,2,3,4 or 2,4,6,8, or 1,3,6,9