<s>[INST] Solve this trigonometric inequality The values of $\lambda$ for which the equation $2\sin (x)-\sqrt \lambda \cos(x)=\sqrt2+\sqrt{2-\lambda}$ has solutions.
MY APPROOACH:
Now I know that max value of LHS of above expression can be $\sqrt{4+\lambda}$ and min value can be $\sqrt{4-\lambda}$. So $\sqrt2+\sqrt{2-\lambda}\in [\sqrt{4-\lambda},\sqrt{4+\lambda}]$. So I get $\lambda \ge \frac{1}{2}$. Now how do I find the max value because I am unable to get a conclusion from $\sqrt{4+\lambda}\ge \sqrt2+\sqrt{2-\lambda}$ 
 [/INST] Obviously, we need $-4\le\lambda\le2$.
\begin{align*}
\sqrt{4+\lambda}&\ge\sqrt{2}+\sqrt{2-\lambda}\\
4+\lambda&\ge2+2\sqrt{2(2-\lambda)}+2-\lambda\\
\lambda&\ge \sqrt{4-2\lambda} \qquad\qquad (\text{Note: this implies that }\lambda\ge0)\\
\lambda^2&\ge4-2\lambda\\
(\lambda+1)^2&\ge5\\
\lambda+1&\ge\sqrt{5}\\
\lambda&\ge\sqrt{5}-1
\end{align*}
So $\sqrt{5}-1\le \lambda\le2$.
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