For \(\tan t=\sqrt3\), \(t=\dfrac\pi3+\pi n\), \(n\in\mathbb Z\). Thus \(3x=\dfrac\pi3+\pi n\), so \(x=\dfrac\pi9+\dfrac{\pi n}{3}\). Answer: C.
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2
Task
One correct answer
Which of the following equations has no real roots?
Solution
The equations \(\sin x=a\) and \(\cos x=a\) have real solutions only when \(|a|\le1\). Tangent and cotangent can take any real value. Since \(\dfrac2{\sqrt3}\gt1\), \(\cos x=\dfrac2{\sqrt3}\) has no real roots. Answer: E.
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3
Task
One correct answer
Solve \(3\cdot\dfrac{\sin x}{\cos x}=\sqrt3\).
Solution
Since \(\dfrac{\sin x}{\cos x}=\tan x\), we get \(\tan x=\dfrac{\sqrt3}{3}\). Hence \(x=\dfrac\pi6+\pi n\), \(n\in\mathbb Z\). Answer: A.
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4
Task
One correct answer
Find the smallest positive root of \(\sin\left(x+\dfrac\pi3\right)=0\).
Solution
\(\sin t=0\) when \(t=\pi n\). Thus \(x=-\dfrac\pi3+\pi n\). For \(n=1\), the smallest positive root is \(x=\dfrac{2\pi}{3}\). Answer: C.
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5
Task
One correct answer
Solve \(2\sin x=1\).
Solution
We have \(\sin x=\dfrac12\). Using \(x=(-1)^n\arcsin a+\pi n\) and \(\arcsin\dfrac12=\dfrac\pi6\), we get \(x=(-1)^n\dfrac\pi6+\pi n\). Answer: E.
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6
Task
One correct answer
Solve \(\tan\dfrac x2=\sqrt3\).
Solution
From \(\tan t=\sqrt3\), \(t=\dfrac\pi3+\pi n\). Thus \(\dfrac x2=\dfrac\pi3+\pi n\), giving \(x=\dfrac{2\pi}{3}+2\pi n\). Answer: D.
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7
Task
One correct answer
If \(2\cos a-5\sin a=0\), then \(\tan a=\)
Solution
From \(2\cos a=5\sin a\), dividing by \(5\cos a\) gives \(\tan a=\dfrac25\). Answer: A.
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8
Task
One correct answer
Find the smallest positive root of \(2\sin x=-1\).
Solution
We have \(\sin x=-\dfrac12\). In \((0,2\pi)\), the solutions are \(\dfrac{7\pi}{6}\) and \(\dfrac{11\pi}{6}\). The smallest positive root is \(\dfrac{7\pi}{6}\). Answer: E.
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9
Task
One correct answer
Solve \(\cos(3x)=\dfrac12\).
Solution
For \(\cos t=\dfrac12\), \(t=\pm\dfrac\pi3+2\pi k\). Thus \(x=\pm\dfrac\pi9+\dfrac{2\pi k}{3}\). Answer: A.
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10
Task
One correct answer
Solve \(\sin x-\sqrt3\cos x=0\).
Solution
From \(\sin x=\sqrt3\cos x\), divide by \(\cos x\) to get \(\tan x=\sqrt3\). Hence \(x=\dfrac\pi3+\pi n\). Answer: D.
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11
Task
One correct answer
Solve \(\tan\left(5x+\dfrac\pi4\right)=\sqrt3\).
Solution
For \(\tan t=\sqrt3\), \(t=\dfrac\pi3+\pi k\). Hence \(5x+\dfrac\pi4=\dfrac\pi3+\pi k\), so \(x=\dfrac\pi{60}+\dfrac{\pi k}{5}\). Answer: B.
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12
Task
One correct answer
Solve \(\cos2x=-\dfrac12\).
Solution
For \(\cos t=-\dfrac12\), \(t=\pm\dfrac{2\pi}{3}+2\pi n\). Thus \(x=\pm\dfrac\pi3+\pi n\). Answer: D.
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13
Task
One correct answer
Find the greatest negative root of \(\cos3x=1\).
Solution
\(\cos t=1\) when \(t=2\pi k\). Thus \(x=\dfrac{2\pi k}{3}\). Taking \(k=-1\) gives the greatest negative root \(-\dfrac{2\pi}{3}\). Answer: C.
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14
Task
One correct answer
Solve \(2\sin^2x+\cos x-1=0\).
Solution
Using \(\sin^2x=1-\cos^2x\), we get \((2\cos x+1)(\cos x-1)=0\). Thus \(\cos x=1\) or \(\cos x=-\dfrac12\), giving \(x=2\pi k\) or \(x=\pm\dfrac{2\pi}{3}+2\pi k\). Answer: D.
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15
Task
Numeric answer
Find the greatest value of the parameter \(a\) for which \[\begin{cases}(2a-1)\sin x+\cos x=2,\\a\sin x+(2a-1)\cos x=a+1\end{cases}\] has infinitely many solutions.
Solution
Let \(u=\cos x\), \(v=\sin x\), so \(u^2+v^2=1\). For \(a\ne1\), the linear system gives \(v=\dfrac3{4a-1}\) and \(u=\dfrac{2a+1}{4a-1}\). Imposing \(u^2+v^2=1\) yields \(4a^2-4a-3=0\), so \(a=1.5\) or \(a=-0.5\). The greatest value is \(1.5\).
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16
Task
Numeric answer
Find the least value of \(a\) for which \[\frac12(\sin x+\sqrt3\cos x)=6-5a-2a^2\] has a solution.
Solution
The left side equals \(\sin\left(x+\dfrac\pi3\right)\). Thus solutions exist exactly when \(|6-5a-2a^2|\le1\). Solving the resulting pair of quadratic inequalities gives the least admissible value \(a=-3.5\).
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17
Task
Numeric answer
Solve \[\begin{cases}5\cos\dfrac{\pi xy}{2}=x^2-8x+21,\\y+5x-4=0.\end{cases}\] If there is a unique solution \((x_0,y_0)\), enter \(x_0+y_0\); otherwise enter the number of solutions.
Solution
We have \(-5\le5\cos\dfrac{\pi xy}{2}\le5\), while \(x^2-8x+21=(x-4)^2+5\ge5\). Equality is possible only when both sides equal \(5\), so \(x=4\). The second equation gives \(y=-16\), and \(\cos(-32\pi)=1\). Thus the unique solution is \((4,-16)\), so \(x_0+y_0=-12\).
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18
Task
Numeric answer
Solve \[\frac{2\cos x+1}{\sqrt{27+6x-x^2}}=0.\] Enter the total number of roots. If there are infinitely many, enter \(100\).
Solution
Because the square root is in the denominator, \(27+6x-x^2\gt0\), so \(x\in(-3,9)\). The numerator is zero when \(\cos x=-\dfrac12\), i.e. \(x=\pm\dfrac{2\pi}{3}+2\pi n\). Exactly four such values lie in \((-3,9)\). Answer: \(4\).
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19
Task
One correct answer
How many roots of \(\cos x=0\) lie in \([0,2\pi]\)?
Solution
\(\cos x=0\) when \(x=\dfrac\pi2+\pi k\). In \([0,2\pi]\), the roots are \(\dfrac\pi2\) and \(\dfrac{3\pi}{2}\), so there are two. Answer: C.
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20
Task
One correct answer
Choose a root of \(\sin4x=-1\).
Solution
\(\sin t=-1\) when \(t=-\dfrac\pi2+2\pi n\). Thus \(x=-\dfrac\pi8+\dfrac{\pi n}{2}\). Taking \(n=1\) gives \(x=\dfrac{3\pi}{8}\). Answer: A.
