Let \(t=2x\). Then \(\sin t\gt-\dfrac{\sqrt2}{2}\), so \(t\in(-\dfrac\pi4+2\pi k;\dfrac{5\pi}4+2\pi k)\). Hence \(x\in(-\dfrac\pi8+\pi k;\dfrac{5\pi}8+\pi k)\). Answer: D.
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2
Task
One correct answer
Find the length of each interval formed by the solutions of \(2\sin x\le1\).
Solution
We have \(\sin x\le\dfrac12\). The boundary points are \(\dfrac\pi6\) and \(\dfrac{5\pi}{6}\). Each solution interval has length \(\dfrac\pi6+2\pi-\dfrac{5\pi}{6}=\dfrac{4\pi}{3}\). Answer: C.
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3
Task
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Solve \(\cos\pi x\gt\dfrac12\).
Solution
Let \(t=\pi x\). Then \(\cos t\gt\dfrac12\) for \(t\in(-\dfrac\pi3+2\pi k;\dfrac\pi3+2\pi k)\). Thus \(x\in(-\dfrac13+2k;\dfrac13+2k)\). Answer: A.
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4
Task
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Find the length of each interval formed by the solutions of \(\cos^2\dfrac x2-\sin^2\dfrac x2\le-\dfrac{\sqrt3}{2}\).
Solution
Using the double-angle identity, the inequality becomes \(\cos x\le-\dfrac{\sqrt3}{2}\). The solution interval is \([\dfrac{5\pi}{6},\dfrac{7\pi}{6}]\), whose length is \(\dfrac\pi3\). Answer: A.
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5
Task
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Solve \(\sin^2x\gt\dfrac14\).
Solution
\(\sin^2x\gt\dfrac14\) is equivalent to \(|\sin x|\gt\dfrac12\). Using the period \(\pi\), \(x\in(\dfrac\pi6+\pi k;\dfrac{5\pi}{6}+\pi k)\). Answer: A.
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6
Task
One correct answer
Solve \(\cos^2x\le\dfrac14\).
Solution
\(\cos^2x\le\dfrac14\) is equivalent to \(|\cos x|\le\dfrac12\). Using period \(\pi\), \(x\in[\dfrac\pi3+\pi k;\dfrac{2\pi}{3}+\pi k]\). Answer: B.
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7
Task
One correct answer
Solve \((2\sin x-3)\tan x\ge0\).
Solution
Since \(2\sin x-3\lt0\) for all \(x\), the product is nonnegative exactly when \(\tan x\le0\). Hence \(x\in(-\dfrac\pi2+\pi n;\pi n]\). Answer: C.
