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?

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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?

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Answer 1 · 2015-08-29T23:52:39

$a = \frac{π}{6}$ $a = pi/6" "$ or $a = 30^{\circ}$ $" "a = 30^@$

Explanation:

You will use three trigonometric identities to solve this equation

$tan x = \frac{sin x}{cos x}$ $tanx = sinx/cosx" "$ , $sec x = \frac{1}{cos x}$ $" "secx = 1/cosx$

and

${cos}^{2} x + {sin}^{2} x = 1$ $cos^2x + sin^2x = 1$

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

$7 \cdot \frac{sin (a)}{cos (a)} + 2 cos (a) = 5 \cdot \frac{1}{cos (a)}$ $7 * sin(a)/cos(a) + 2 cos(a) = 5 * 1/cos(a)$

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

For the interval $[0, π]$ $[0, pi]$ , $cos (a) = 0$ $cos(a) = 0$ for $a = \frac{π}{2}$ $a = pi/2$ .

So, multiply $2 cos (a)$ $2cos(a)$ by $1 = \frac{cos (a)}{cos (a)}$ $1 = cos(a)/cos(a)$ to get rid of the denominators

$7 \cdot \frac{sin (a)}{cos (a)} + 2 \frac{{cos}^{2} (a)}{cos (a)} = 5 \cdot \frac{1}{cos (a)}$ $7 * sin(a)/cos(a) + 2 cos^2(a)/cos(a) = 5 * 1/cos(a)$

This is equivalent to

$7 \cdot sin (a) + 2 {cos}^{2} (a) - 5 = 0$ $7 * sin(a) + 2cos^2(a) - 5 = 0$

Use ${cos}^{2} (a) = 1 - {sin}^{2} (a)$ $cos^2(a) = 1 - sin^2(a)$ to get

$7 \cdot sin (a) + 2 [1 - {sin}^{2} (a)] - 5 = 0$ $7 * sin(a) + 2[1-sin^2(a)] - 5 = 0$

The quadratic form of this equaation will thus be

$- 2 {sin}^{2} (a) + 7 sin (a) - 3 = 0$ $-2sin^2(a) + 7sin(a) - 3= 0$

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

${sin}_{1, 2} (a) = \frac{- 7 \pm \sqrt{7^{2} - 4 \cdot (- 2) \cdot (- 3)}}{2 \cdot (- 2)}$ $sin_(1,2)(a) = (-7 +- sqrt(7^2 - 4 * (-2) * (-3)))/(2 * (-2))$

${sin}_{1, 2} (a) = \frac{- 7 \pm \sqrt{25}}{(- 4)} = \frac{- 7 \pm 5}{(- 4)}$ $sin_(1,2)(a) = (-7 +- sqrt(25))/((-4)) = (-7 +- 5)/((-4))$

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

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

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

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

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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

Explanation:

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