Given that 7tan a + 2 cos a= 5sec a,derive a quadratic equation for sin a. Hence, find all values of a in the interval [0,pi] which satisfy the given equation?

1 Answer
Aug 29, 2015

Answer:

#a = pi/6" "# or #" "a = 30^@#

Explanation:

You will use three trigonometric identities to solve this equation

#tanx = sinx/cosx" "#, #" "secx = 1/cosx#

and

#cos^2x + sin^2x = 1#

Start by replacing #tan(a)# and #sec(a)# into the original equation

#7 * sin(a)/cos(a) + 2 cos(a) = 5 * 1/cos(a)#

Right from the start, any value of #a# that would make #cos(a) = 0# will be excluded from the solutions set.

For the interval #[0, pi]#, #cos(a) = 0# for #a = pi/2#.

So, multiply #2cos(a)# by #1 = cos(a)/cos(a)# to get rid of the denominators

#7 * sin(a)/cos(a) + 2 cos^2(a)/cos(a) = 5 * 1/cos(a)#

This is equivalent to

#7 * sin(a) + 2cos^2(a) - 5 = 0#

Use #cos^2(a) = 1 - sin^2(a)# to get

#7 * sin(a) + 2[1-sin^2(a)] - 5 = 0#

The quadratic form of this equaation will thus be

#-2sin^2(a) + 7sin(a) - 3= 0#

Use the quadratic formula to get the two roots of the quadratic

#sin_(1,2)(a) = (-7 +- sqrt(7^2 - 4 * (-2) * (-3)))/(2 * (-2))#

#sin_(1,2)(a) = (-7 +- sqrt(25))/((-4)) = (-7 +- 5)/((-4))#

The two roos will thus be

#color(red)(cancel(color(black)(sin_1(a) = (-7-5)/((-4)) = 3)))" "# and #" "sin_2(a)= (-7 + 5)/((-4)) = 1/2" "color(green)(sqrt())#

Since #sin(a) = 3# is outside the range of the sine function, the only valid solution will be #sin(a) = 1/2#.

The value of #a# that corresponds to this value is

#a = color(green)(pi/6)" "# or #" "a = color(green)(30^@)#

Do a quick check to make sure that the calculations are correct

#7 * tan(pi/6) + 2 * cos(pi/6) = 5 * sec(pi/6)#

#7 * sqrt(3)/3 + 2 * sqrt(3)/2 = 5 * 2/sqrt(3)#

#14sqrt(3) + 6sqrt(3) = 20sqrt(3)" "color(green)(sqrt())#