Question

Solve the following trigonometric equation for general solution :

2cos^2 theta+ 3sin theta = 0​

Answer 1

Step-by-step explanation:

2cos² theta + 3 sin theta =0

2(1-sin²theta) + 3 sin theta = 0

2 sin² theta - 3 sin theta -2 =0

(2sin theta +1)(sin theta -2) = 0

sin theta=/= 2

sin theta = -1/2

theta= nPie + (-1)ⁿ(-pie/6)

Answer 2

GiveN :-

• A trignometric equation is given to us .

$\sf 2cos^2\theta+3 sin\theta = 0$

To FinD:-

• The general solution of equation .

SolutioN :-

We need to find the general solution of the given trignometric equation. We know that if $\sf \alpha$ is the value of $\theta$ , then the general solution of the equation will be ,

$\sf:\implies\pink{ \theta = n\pi + (-1)^n(\alpha) }$

$\red{\bigstar}\underline{\textsf{So , let's solve out the equation , }}$

$\sf:\implies 2cos^2\theta+3 sin\theta = 0 \\\\\sf:\implies 2(1-sin^2\theta) + 3sin\theta = 0 \\\\\sf:\implies 2- 2 sin^2\theta + 3 sin\theta = 0 \\\\\sf:\implies 2sin^2\theta - 3sin\theta-2=0\\\\\sf:\implies sin\theta = \dfrac{-(-3)\pm\sqrt{(-3)^2-4(2)(-2)}}{2(2)} \\\\:\sf\implies sin\theta = \dfrac{3\pm \sqrt{9+16}}{4} \qquad \bigg\lgroup \red{\tt Using \ Quadratic \ Formula }\bigg\rgroup \\\\\sf:\implies sin\theta = \dfrac{3\pm \sqrt{25}}{4}\\\\\sf:\implies sin\theta = \dfrac{3+5}{4},\dfrac{3-5}{4} \\\\\sf:\implies sin\theta = 2 , \dfrac{-1}{2}$

Here 2 is not possible value of 2 because the range of sin is from $\sf -1\leq \theta \leq 1$.

$\sf:\implies sin\theta = -\dfrac{1}{2}\\\\\sf:\implies sin\theta = sin\bigg(-\dfrac{\pi}{6}\bigg) \\\\\sf:\implies \boxed{\pink{\sf{ \theta =-\dfrac{\pi}{6} }}}$

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$\red{\bigstar}\underline{\textsf{Therefore the general Solution will be , }}$

$\sf:\implies \theta_{\alpha}= n\pi + (-1)^n(\alpha) \\\\\sf:\implies \theta = n\pi + (-1)^n \bigg(-\dfrac{\pi}{6}\bigg) \\\\\sf:\implies\underset{\blue{\sf General \ Solution }}{\underbrace{\boxed{\pink{\frak {\theta = n\pi -(-1)^n \bigg(\dfrac{\pi}{6}\bigg) }}}}}$