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subhan2010 · Jul 24, 2012, 06:28 AM

QUESTION:the general solution of equation (√3 - 1)sin∅+(√3 + 1)c0s∅=2 is
1) ∅=2nπ+- (π/4)+(π/12)
2) ∅= nπ + (-1)^π (π/4)+(π/12)
3) ∅=2(n+1)π (π/4)+(π/12)

The answer is option 1 , I don't know why the last 2 options can be the solutions of the question as well.my teacher told me that to use 2nπ with sin and cos functions and nπ with tan functions.can someone please describe in detail about the periods used in solutions of trig.equations.THANKS

ebaines · Jul 24, 2012, 10:11 AM

Remember that the sine and cosine functions have periods of $2 \pi$ , so that $\sin(\theta) = \sin (\theta \pm 2n \pi)$ and $\cos(\theta) = \cos(\theta \pm 2n \pi)$ . Any solution to an equation involving sines and/or cosines will have general solutions with $\pm 2 n \pi$ . Also recall that $\sin(\theta) = - \sin (\theta \pm \pi)$ , and similarly $\cos(\theta) = - \cos (\theta \pm \pi )$ , and hence the tangent function is periodic with period $\pi$ , since the negative signs cancel out:

$<br /> \tan \theta = \frac {\sin (\theta \pm n \pi)}{\cos (\theta \pm n \pi)} = \tan (\theta \pm n \pi)$ .

By the way, you wrote that the solution is

$\phi = 2 n \pi \pm \frac n 4 + \frac n {12}$

but I believe it's actually $\phi = \frac n 4 + \frac n {12} \pm 2 n \pi$

subhan2010 · Jul 24, 2012, 09:32 PM

Thank you..