Answer:To solve the equation sin(x) + 1 = cos(x), follow these steps:
Rearrange the equation to isolate the trigonometric functions on one side: [ \sin(x) = \cos(x) - 1 ]
Use trigonometric identities to express (\sin(x)) and (\cos(x)) in terms of other trigonometric functions: [ \sin(x) = \frac{2\tan\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)} ] [ \cos(x) = \frac{1 - \tan^2\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)} ]
Substitute these expressions back into the equation: [ \frac{2\tan\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)} = \frac{1 - \tan^2\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)} - 1 ]
Simplify the right-hand side: [ \frac{2\tan\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)} = \frac{1 - \tan^2\left(\frac{x}{2}\right) - (1 + \tan^2\left(\frac{x}{2}\right))}{1 + \tan^2\left(\frac{x}{2}\right)} ] [ \frac{2\tan\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)} = \frac{-2\tan^2\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)} ]
Cross-multiply: [ 2\tan\left(\frac{x}{2}\right) = -2\tan^2\left(\frac{x}{2}\right) ]
Divide both sides by 2: [ \tan\left(\frac{x}{2}\right) = -\tan^2\left(\frac{x}{2}\right) ]
Rearrange: [ \tan^2\left(\frac{x}{2}\right) + \tan\left(\frac{x}{2}\right) = 0 ]
Factor: [ \tan\left(\frac{x}{2}\right)\left(\tan\left(\frac{x}{2}\right) + 1\right) = 0 ]
Apply the zero product property:
Set each factor equal to zero: [ \tan\left(\frac{x}{2}\right) = 0 \quad \text{or} \quad \tan\left(\frac{x}{2}\right) + 1 = 0 ]
Solve for (x):
For (\tan\left(\frac{x}{2}\right) = 0), we get (x = 2k\pi) where (k) is an integer.
For (\tan\left(\frac{x}{2}\right) + 1 = 0), we get (\tan\left(\frac{x}{2}\right) = -1), which gives (x = (2k + 1)\pi) where (k) is an integer.
Therefore, the solutions for (x) are (x = 2k\pi) and (x = (2k + 1)\pi), where (k) is an integer.
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