Exam 6: Inverse Circular Functions and Trigonometric Equations

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Solve the equation for solutions in the interval [0, 2 [0,2π).[ 0,2 \pi ) . - sinxcosx=12\sin x \cos x = \frac { 1 } { 2 }

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Solve the problem. -A painting 1 meter high and 3 meters from the floor will cut off an angle θ\theta to an observer, where θ=tan1(xx2+1.6)\theta = \tan ^ { - 1 } \left( \frac { x } { x ^ { 2 } + 1.6 } \right) , assuming that the observer is xx feet from the wall where the painting is displayed and that the eyes of the observer are 1.61.6 meters above the ground (see the figure). Find the value of θ\theta for x=3x = 3 . Round to the nearest tenth of a degree.  Solve the problem. -A painting 1 meter high and 3 meters from the floor will cut off an angle  \theta  to an observer, where  \theta = \tan ^ { - 1 } \left( \frac { x } { x ^ { 2 } + 1.6 } \right) , assuming that the observer is  x  feet from the wall where the painting is displayed and that the eyes of the observer are  1.6  meters above the ground (see the figure). Find the value of  \theta  for  x = 3 . Round to the nearest tenth of a degree.

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Give the exact value of the expression. - cos(2arcsin14)\cos \left( 2 \arcsin \frac { 1 } { 4 } \right)

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Use the parallelogram rule to find the magnitude of the resultant force for the two forces shown in the figure. Round to one decimal place. -Explain what is wrong with the following solution for the equation cos2θ=3\cos 2 \theta = \sqrt { 3 } in the interval [0,2π)[ 0,2 \pi ) . cos2θ=3\cos 2 \theta = \sqrt { 3 } cosθ=32\cos \theta = \frac { \sqrt { 3 } } { 2 } θ=π6 or θ=11π6\theta = \frac { \pi } { 6 } \text { or } \theta = \frac { 11 \pi } { 6 }

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Solve the equation for exact solutions. - 6arcsinx=π6 \arcsin x = \pi

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Use a calculator to give the real number value. Round the answer to 7 decimal places. - y=sin1(0.2079)y = \sin ^ { - 1 } ( 0.2079 )

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Give the exact value of the expression. - cos(arcsin513+arccos35)\cos \left( \arcsin \frac { 5 } { 13 } + \arccos \frac { 3 } { 5 } \right)

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Solve the problem. -The formula for the up-and-down motion of a weight on a spring is given by S=asin(km)t\mathrm { S } = \mathrm { a } \sin \left( \frac { \sqrt { \mathrm { k } } } { \mathrm { m } } \right) \mathrm { t } where a is the radius of the circle, k\mathrm { k } is the spring constant, m\mathrm { m } is the mass, and t\mathrm { t } is the time. Solve the equation for t.

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Use a calculator to give the real number value. Round the answer to 7 decimal places. - y=cos1(0.3907)y = \cos ^ { - 1 } ( - 0.3907 )

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Use a calculator to find the value. Give answers as real numbers and round to 4 decimal places, if necessary. - sin(cos10.8324)\sin \left( \cos ^ { - 1 } 0.8324 \right)

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Solve the equation for x, where x is restricted to the given interval. - y=3tan2x1, for x in (π4,π4)y = 3 \tan 2 x - 1 \text {, for } x \text { in } \left( - \frac { \pi } { 4 } , \frac { \pi } { 4 } \right)

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Give the degree measure of . -ϴ = arcsin (1)

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Solve. -The output voltage for an AC\mathrm { AC } generator is approximated by v=156cos(120πtπ3)\mathrm { v } = 156 \cos \left( 120 \pi \mathrm { t } - \frac { \pi } { 3 } \right) . Find the smallest positive value of t\mathrm { t } for which the output is 75 volts. Round values to 4 decimal places.

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Solve the problem. -The position of a weight on a spring relative to the point of equilibrium is given by y = 4 cos 6t - 2 sin 6t, Where the arguments are in radians and t is in seconds. Find the smallest value of t for which the Weight is at the point of equilibrium (y = 0). Give your answer to the nearest hundredth.

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Solve the equation for exact solutions over the interval [0, 2 [0,2π)[ 0,2 \pi ) - csc5x4cscx=0\csc ^ { 5 } x - 4 \csc x = 0

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Solve the equation for exact solutions. - sin1x+tan1x=0\sin ^ { - 1 } x + \tan ^ { - 1 } x = 0

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Find the exact value of the real number y. - y=csc1(1)y = \csc ^ { - 1 } ( - 1 )

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Solve the equation for exact solutions over the interval [0, 2 [0,2π)[ 0,2 \pi ) - sec2x2=tan2x\sec ^ { 2 } x - 2 = \tan ^ { 2 } x

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Solve the problem. -The figure shows a stationary spy satellite positioned 12,000 miles above the equator. What percent, to the nearest tenth, of the equator can be seen from the satellite? The diameter of Earth is 7927 Miles at the equator. Solve the problem. -The figure shows a stationary spy satellite positioned 12,000 miles above the equator. What percent, to the nearest tenth, of the equator can be seen from the satellite? The diameter of Earth is 7927 Miles at the equator.

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