Question

If (sin²θ - 3sinθ + 2)/cos²θ = 1, then θ = ?​

Answer 1

Answer:

$\rm θ = 90^{\circ} \: \: or \: \: 30^{\circ}$

Step-by-step explanation:

We have,

$\rm \dfrac{( \sin ^2 θ -3 \sin θ +2)}{\cos ^2 θ}=1$

$\rm \dfrac{( \sin ^2 θ -3 \sin θ +2)}{1-\sin ^2 θ}=1$

$\rm \sin ^2 θ -3 \sin θ +2=1-\sin ^2 θ$

$\rm 2 \sin ^2 θ -3 \sin θ +1=0$

let $\rm \sin \theta =x$

$\rm 2 x^2 -3 x +1=0$

$\rm x = \dfrac{-b ± \sqrt{b^2 -4ac}}{2a}$

$\rm x= \dfrac{-(-3) ± \sqrt{(-3)^2 -4(2)(1)}}{2(2)}$

$\rm x= \dfrac{3± \sqrt{9 -8}}{4}$

$\rm x= \dfrac{3± \sqrt{1}}{4}$

$\rm x= \dfrac{3± 1}{4}$

$\rm x= \dfrac{4}{4} \: \: or \: \: \dfrac{2}{4}$

$\rm x= 1 \: or \: \dfrac{1}{2}$

$\rm \sin \theta = 1 \: \: or \: \: \dfrac{1}{2}$

$\rm \sin \theta = \sin 90^{\circ} \: \: or \: \: \sin 30^{\circ}$

$\rm θ = 90^{\circ} \: \: or \: \: 30^{\circ}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given Trigonometric equation is

$\rm \: \dfrac{ {sin}^{2}\theta - 3sin\theta + 2 }{ {cos}^{2}\theta } = 1$

$\rm \: {sin}^{2}\theta - 3sin\theta + 2 = {cos}^{2}\theta$

We know,

$\boxed{\tt{ \: \: {sin}^{2}x + {cos}^{2}x = 1 \: \: }}$

So, using this identity, we get

$\rm \: {sin}^{2}\theta - 3sin\theta + 2 = 1 - {sin}^{2}\theta$

$\rm \: {sin}^{2}\theta - 3sin\theta + 2 - 1 + {sin}^{2}\theta = 0$

$\rm \: 2{sin}^{2}\theta - 3sin\theta + 1 = 0$

So, on splitting the middle terms, we get

$\rm \: 2{sin}^{2}\theta - 2sin\theta - sin\theta + 1 = 0$

$\rm \: 2sin\theta (sin\theta - 1) - 1(sin\theta - 1) = 0$

$\rm \: (2sin\theta - 1)(sin\theta - 1) = 0$

$\rm\implies \:sin\theta = \dfrac{1}{2} \: \: or \: \: sin\theta = 1$

Consider, Case :- 1

$\rm \: sin\theta = \dfrac{1}{2}$

$\rm \: sin\theta = sin\dfrac{\pi}{6}$

We know,

$\boxed{\tt{ sinx = siny \rm\implies \: \sf x = n\pi + {( - 1)}^{n}y \: \forall \: n \in \: Z}}$

So, using this result, we get

$\rm\implies \:\theta = n\pi + {( - 1)}^{n}\dfrac{\pi}{6} \: \: \: \forall \: n \: \in \: Z$

Consider Case :- 2

$\rm \: sin\theta = 1$

$\rm \: sin\theta = sin\dfrac{\pi}{2}$

$\rm\implies \:\theta = n\pi + {( - 1)}^{n}\dfrac{\pi}{2} \: \: \: \forall \: n \: \in \: Z$

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More to know

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf T-eq & \bf Solution \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf sinx = 0 & \sf x = n\pi \: \forall \: n \in \: Z\\ \\ \sf cosx = 0 & \sf x = (2n + 1)\dfrac{\pi}{2}\: \forall \: n \in \: Z\\ \\ \sf tanx = 0 & \sf x = n\pi\: \forall \: n \in \: Z\\ \\ \sf sinx = siny & \sf x = n\pi + {( - 1)}^{n}y \: \forall \: n \in \: Z\\ \\ \sf cosx = cosy & \sf x = 2n\pi \pm \: y\: \forall \: n \in \: Z\\ \\ \sf tanx = tany & \sf x = n\pi + y \: \forall \: n \in \: Z\end{array}} \\ \end{gathered}\end{gathered}$