A

9000046502

Level: 
A
Identify the optimal first step convenient to solve the following trigonometric equation. Do not consider the step which is possible but does not help to solve the equation. \[ \cos 3x = 0.5 \]
substitution \( 3x = z\)
substitution \( \cos x = z\)
\(\cos ^{3}x -\sin ^{3}x = 0.5\)
\(\cos x = \frac{0.5} {3} \)

9000046503

Level: 
A
Identify the optimal first step convenient to solve the following trigonometric equation. Do not consider the step which is possible but does not help to solve the equation. \[ \mathop{\mathrm{tg}}\nolimits \left (-x + \frac{\pi } {6}\right ) = \sqrt{3} \]
substitution \( - x + \frac{\pi } {6} = z\)
\(\mathop{\mathrm{tg}}\nolimits (-x) = \sqrt{3} - \frac{\pi } {6}\)
\(\mathop{\mathrm{tg}}\nolimits ^{2}\left (-x + \frac{\pi } {6}\right ) = 3\)
\(\frac{\sin \left (-x+ \frac{\pi }{6} \right )} {\cos \left (-x+ \frac{\pi }{6} \right )} = \sqrt{3}\)

9000046504

Level: 
A
Identify the optimal first step convenient to solve the following trigonometric equation. Do not consider the step which is possible but does not help to solve the equation. \[ \cos \left (x + \frac{\pi } {3}\right ) = \frac{\sqrt{3}} {2} \]
substitution \( x + \frac{\pi } {3} = z\)
\(\cos ^{2}\left (x + \frac{\pi } {3}\right ) = \frac{3} {4}\)
substitution \( \frac{\sqrt{3}} {2} = z\)
\(\cos x\cdot \cos \frac{\pi }{3} -\sin x\cdot \sin \frac{\pi }{3} = \frac{\sqrt{3}} {2} \)

9000046510

Level: 
A
Identify the optimal first step convenient to solve the following trigonometric equation. Do not consider the step which is possible but does not help to solve the equation. \[ 2\sin ^{2}x -\sin x - 1 = 0 \]
substitution \( \sin x = z\)
substitution \( \sin ^{2}x = z\)
\(2\sin ^{2}x -\sin x = 1\)
\(2\sin ^{2}x -\sin x =\sin ^{2}x +\cos ^{2}x\)

9000045709

Level: 
A
Let \(\omega \) be the angle between the solid diagonal of a box and the base of this box. Find the expression which allows to find \(\omega \).
\(\mathop{\mathrm{tg}}\nolimits \omega = \frac{\sqrt{2}} {2} \)
\(\cos \omega = \frac{\sqrt{2}} {2} \)
\(\sin \omega = \frac{\sqrt{2}} {2} \)
\(\mathop{\mathrm{cotg}}\nolimits \omega = \frac{\sqrt{2}} {2} \)