Question

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?

Answer 1

`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())`