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

User LifeQuery
by
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1 Answer

11 votes

Answer:

π

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

User Eme Eme
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