Solve the equation tan^2 theta =((v3)/ 6 )sec theta on the
interval 0 less than or equal to theta which less than 2pi.

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Ответ:

alsafy5124

23.09.2021

Mathematics

First write the equations in the slope-intercept form:
$y=mx+b \\ \\
m=\frac{y_2-y_1}{x_2-x_1}$

Line 1 passes through the points (-6,-6) and (-4,-3).
$m=\frac{-3-(-6)}{-4-(-6)}=\frac{-3+6}{-4+6}=\frac{3}{2} \\ \\
(-6,-6) \to x=-6, \ y=-6 \\
-6=\frac{3}{2} \times (-6) + b \\
-6=-9+b \\
b=3 \\ \\ y=\frac{3}{2}x+3$

Line 2 passes through the points (0,3) and (2,6).
$m=\frac{6-3}{2-0}=\frac{3}{2} \\ \\
(0,3) \to x=0, \ y=3 \\
3=\frac{3}{2} \times 0 + b \\
b=3 \\ \\
y=\frac{3}{2}x+3$

These are two identical lines so the system of equations has infinitely many solutions. The answer is D.

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Solve the equation tan^2 theta =((v3)/ 6 )sec theta on the interval 0 less than or equal to theta which less than 2pi.