Solve for $\displaystyle x,y$: $\displaystyle \left(\sin(x-y)+1\right)\left(2\cos(2x-y)+1\right)=6$.
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Originally Posted by james_bond Solve for $\displaystyle x,y$: $\displaystyle \left(\sin(x-y)+1\right)\left(2\cos(2x-y)+1\right)=6$. You just have to remember that $\displaystyle -1 \leq \sin \alpha \leq 1$ and $\displaystyle -1 \leq \cos \alpha \leq 1$ for every $\displaystyle \alpha$ real
It just seemed too difficult but yeah thanks (again).
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