Can you solve 9sin^2theta - 6sintheta = 1 over [0 degree, 360 degrees) ?

Question

Aaliyah Abdullah · Accepted Answer

#theta=53.58 color(white)(888)# and #color(white)(888)theta=126.42#

#theta=352.06 color(white)(888)# and # color(white)(888)theta=187.93 color(white)(888)# ( 2 d.p. )

#9sin^2(theta)-6sin(theta)=1# #9sin^2(theta)-6sin(theta)-1=0# This is just a quadratic equation, which we can easily solve: Let # \ \ \ \u=sin(theta)# Then: #9u^2-6u-1=0# Using quadratic formula: #u=(-b+-sqrt(b^2-4ac))/(2a)# #u=(-(-6)+-sqrt((-6)^2-(4(9)(-1))))/((2)(9))# #u=(6+-sqrt(72))/(18)# #u=(1+-sqrt(2))/3# But # \ \ \ \ u=sin(theta)# #sin(theta)=(1+sqrt(2))/3# #sin(theta)=(1-sqrt(2))/3# #theta=arcsin(sin(theta))=arcsin((1+sqrt(2))/3)# #theta=arcsin(sin(theta))=arcsin((1-sqrt(2))/3)# For: #[0,360]# #theta=arcsin((1+sqrt(2))/3)# and #theta=180-arcsin((1+sqrt(2))/3)# #theta=360+arcsin((1-sqrt(2))/3)# and #theta=180-arcsin((1-sqrt(2))/3)# Approximate values: #theta=53.58 color(white)(888)# and #color(white)(888)theta=126.42# #theta=352.06 color(white)(888)# and # color(white)(888)theta=187.93 color(white)(888)# ( 2 d.p. )

Elijah Alexander · Answer

undefined To solve the equation $9\sin^2(\theta) - 6\sin(\theta) = 1$ over the interval $[0^\circ, 360^\circ)$, we can first rewrite it as a quadratic equation in terms of $\sin(\theta)$:

$$9\sin^2(\theta) - 6\sin(\theta) - 1 = 0$$

Now, let $u = \sin(\theta)$, then the equation becomes:

$$9u^2 - 6u - 1 = 0$$

This is a quadratic equation, which we can solve using the quadratic formula:

$$u = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$$

where $a = 9$, $b = -6$, and $c = -1$.

Plugging in the values:

$$u = \frac{{-(-6) \pm \sqrt{{(-6)^2 - 4(9)(-1)}}}}{{2(9)}}$$

$$u = \frac{{6 \pm \sqrt{{36 + 36}}}}{{18}}$$

$$u = \frac{{6 \pm \sqrt{72}}}{{18}}$$

$$u = \frac{{6 \pm 6\sqrt{2}}}{{18}}$$

$$u = \frac{{1 \pm \sqrt{2}}}{{3}}$$

Now, we have two possible solutions for $u$:

1. $u = \frac{{1 + \sqrt{2}}}{{3}}$
2. $u = \frac{{1 - \sqrt{2}}}{{3}}$

Since $\sin(\theta) = u$, we need to find the corresponding values of $\theta$ within the given interval $[0^\circ, 360^\circ)$.

For $u = \frac{{1 + \sqrt{2}}}{{3}}$:
$$\theta = \sin^{-1}\left(\frac{{1 + \sqrt{2}}}{{3}}\right)$$

For $u = \frac{{1 - \sqrt{2}}}{{3}}$:
$$\theta = \sin^{-1}\left(\frac{{1 - \sqrt{2}}}{{3}}\right)$$

Once we have obtained the values of $\theta$, we need to ensure they fall within the interval $[0^\circ, 360^\circ)$.