We verify this value is attainable. Suppose $ \sin 3x = 1 $ and $ \sin x = 1 $. Then $ x = rac{\pi}{2} + 2\pi n $. Check $ \sin 3x = \sin\left(rac{3\pi}{2}

February 22, 2026

$We verify this value is attainable. Suppose $ \sin 3x = 1 $ and $ \sin x = 1 $. Then $ x = rac{\pi}{2} + 2\pi n $. Check $ \sin 3x = \sin\left( rac{3\pi}{2}$

f(x) = 1 + rac{1}{2}(-2 \sin 3x \sin x) = 1 - \sin 3x \sin x
We now seek the minimum value of $ f(x) = 1 - \sin 3x \sin x $. Since $ -1 \leq \sin 3x \sin x \leq 1 $, the maximum of $ \sin 3x \sin x $ is 1, so the minimum of $ f(x) $ is:
- 1 = 0
ight) = -1
e 1 $. Not simultaneous. Try to find $ x $ such that $ \sin 3x \sin x = 1 $. This requires both $ |\sin 3x| = 1 $ and $ |\sin x| = 1 $, and same sign. But suppose $ \sin x = 1 \Rightarrow x = rac{\pi}{2} $, then $ \sin 3x = \sin\left(rac{3\pi}{2}
ight) = -1 $. Then $ \sin 3x \sin x = -1 $.

We verify this value is attainable. Suppose $ \sin 3x = 1 $ and $ \sin x = 1 $. Then $ x = rac{\pi}{2} + 2\pi n $. Check $ \sin 3x = \sin\left(rac{3\pi}{2}

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