# Trigonometric equation:

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• Dec 15th 2008, 08:16 PM
cmr7
Trigonometric equation:
I am trying to find the solutions for this equation.
I have worked it twice and keep getting different answers. Can someone show me step by step how to work it?
sin(2x)-cos(x)=0
• Dec 15th 2008, 08:19 PM
Chris L T521
Quote:

Originally Posted by cmr7
I am trying to find the solutions for this equation.
I have worked it twice and keep getting different answers. Can someone show me step by step how to work it?
sin(2x)-cos(x)=0

Recall that $\displaystyle \sin\left(2x\right)=2\sin\left(x\right)\cos\left(x \right)$

Thus, $\displaystyle \sin\left(2x\right)-\cos\left(x\right)=0\implies 2\sin\left(x\right)\cos\left(x\right)-\cos\left(x\right)=0\implies \cos\left(x\right)\left[2\sin\left(x\right)-1\right]=0$

Thus, either $\displaystyle \cos\left(x\right)=0$ or $\displaystyle 2\sin\left(x\right)-1=0\implies \sin\left(x\right)=\tfrac{1}{2}$

Can you take it from here? Also, is there a restriction on x [i.e. $\displaystyle 0\leq x\leq2\pi$]??
• Dec 15th 2008, 08:21 PM
cmr7
I think my brain is just fried and I am over thinking things tonight. No restrictions on x. Thank you for all for your help!