Question

Find the general solution of the equation: 2 sin² θ = 3 cos θ.

Answer 1

To find the general Solution
of the equation , convert the equation in either sine or cosine

$2 \: {sin}^{2} \theta = 3 \: cos \: \theta \\ \\ 2(1 - {cos}^{2}\theta) = 3 \: cos \: \theta \\ \\ - 2 {cos}^{2} \theta - 3 \: cos \: \theta + 2 = 0 \\ \\ 2 {cos}^{2} \theta + 3 \: cos \: \theta - 2 = 0 \\ \\2 {cos}^{2} \theta + 4 \: cos \: \theta - cos \: \theta - 2 = 0 \\ \\2 \: cos \: \theta(cos \: \theta + 2) - 1(cos \: \theta + 2) = 0 \\ \\ (cos \: \theta + 2)(2 \: cos \: \theta - 1) = 0$
So

$cos \: \theta + 2 = 0 \\ \\ cos \: \theta = - 2$
discard this value because cosine can't reach till -2.
$(2 \: cos \: \theta - 1) = 0 \\ \\ 2 \: cos \: \theta = 1 \\ \\ cos \: \theta = \frac{1}{2} \\ \\ \theta = {cos}^{ - 1} ( \frac{1}{2} ) \\ \\ \theta = {cos}^{ - 1}(cos \: \frac{\pi}{3} ) \\ \\ \theta = \frac{\pi}{3}$

General Solution of the equation is

$\theta = ±\frac{\pi}{3} + 2\pi \: k$
where k is any integer

Answer 2

Solution :

Here I am using A instead of theta.

2sin²A = 3cosA

=> 2( 1 - cos²A ) - 3cosA = 0

=> 2 - 2cos²A - 3cosA = 0

=> 2cos²A + 3cosA - 2 = 0

Splitting the middle term,

=> 2cos²A - cosA + 4cosA - 2 = 0

=>cosA( 2cosA - 1 )+2( 2cosA - 1 )=0

=> ( 2cosA - 1 )( cosA + 2 ) = 0

Therefore ,

2cosA - 1 = 0 or cosA + 2 = 0

cosA = 1/2 or cosA = -2 ( not possible )

[ Since , cos²A ≥ -1 ]


Now ,

cos A = 1/2

A = 2nπ ± ( π/3 )

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