解一道三角不等式

解不等式：$\sin x\cdot \sin 7x>\dfrac 14$．

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解　$\left(k\pi+\dfrac{\pi}{12},k\pi+\dfrac{\pi}{10}\right)\cup\left(k\pi+\dfrac{3\pi}{10},k\pi+\dfrac{5\pi}{12}\right)\cup\left(k\pi+\dfrac{7\pi}{12},k\pi+\dfrac{7\pi}{10}\right)\cup\left(k\pi+\dfrac{9\pi}{10},k\pi+\dfrac{11\pi}{12}\right)$．

由$7$倍角公式

$$\sin 7x=7\sin x-56\sin^3x+112\sin^5x-64\sin^7x$$ 可得原不等式等价于 $$256\sin^8x-448\sin^6x+224\sin^4x-28\sin^2x+1<0.$$ 令$t=2\sin x$，则原不等式等价于 $$t^8-7t^6+14t^4-7t^2+1<0,$$ 也即 $$\left(t^2+\dfrac{1}{t^2}\right)^2-7\left(t^2+\dfrac{1}{t^2}\right)+12<0,$$ 即 $$3<t^2+\dfrac{1}{t^2}<4,$$ 解得 $$\dfrac{\sqrt 6-\sqrt 2}2<|t|<\dfrac{\sqrt 5-1}2\lor\dfrac{\sqrt 5+1}2<|t|<\dfrac{\sqrt 6+\sqrt 2}2,$$ 即 $$\dfrac{\sqrt 6-\sqrt 2}4<|\sin x|<\dfrac {\sqrt 5-1}4\lor\dfrac{\sqrt 5+1}4<|\sin x|<\dfrac{\sqrt 6+\sqrt 2}4,$$ 故所求解集为 $$\left(k\pi+\dfrac{\pi}{12},k\pi+\dfrac{\pi}{10}\right)\cup\left(k\pi+\dfrac{3\pi}{10},k\pi+\dfrac{5\pi}{12}\right)\cup\left(k\pi+\dfrac{7\pi}{12},k\pi+\dfrac{7\pi}{10}\right)\cup\left(k\pi+\dfrac{9\pi}{10},k\pi+\dfrac{11\pi}{12}\right).$$

注　联想

$$\dfrac 14=\sin\dfrac{\pi}{12}\cdot \cos\dfrac{\pi}{12}=\cos\dfrac{\pi}5\cdot \cos\dfrac{2\pi}5,$$ 也即 $$\dfrac 14=\sin\dfrac{\pi}{12}\cdot \sin\dfrac{7\pi}{12}=\sin \dfrac{3\pi}{10}\cdot \sin\dfrac{21\pi}{10},$$ 结合$f(x)=\sin x\cdot \sin 7x$的图象即得．