Question

The solution set of the equation 4sin θ cos θ - 2cos θ - 2√3sin θ + √3 = 0 in the interval (0,2π) is​

Answer 1

SOLUTION

TO DETERMINE

The solution set of the equation

4sin θ cos θ - 2cos θ - 2√3sin θ + √3 = 0

in the interval (0,2π)

EVALUATION

Here the given equation is

4sin θ cos θ - 2cos θ - 2√3sin θ + √3 = 0

We now solve for θ as below

4sin θ cos θ - 2cos θ - 2√3sin θ + √3 = 0

⇒ 2cos θ ( 2sin θ - 1 ) - √3 ( 2sin θ - 1 ) = 0

⇒ ( 2sin θ - 1 ) ( 2cos θ - √3 ) = 0

Now 2sin θ - 1 = 0 gives

$\displaystyle\sf{2 \sin \theta = 1}$

$\displaystyle\sf{ \implies \: \sin \theta = \frac{1}{2} }$

$\displaystyle\sf{ \implies \: \sin \theta = \sin \: \frac{\pi}{6} }$

$\displaystyle\sf{ \implies \: \theta = \: \frac{\pi}{6} \: \: and \: \: \frac{5\pi}{6} \: \: \big( \because \: \theta \: \in \: (0,2\pi) \big) }$

Again ( 2cos θ - √3 ) = 0 gives

$\displaystyle\sf{ \: 2 \cos \theta = \sqrt{3} }$

$\displaystyle\sf{ \implies \: \cos \theta = \frac{ \sqrt{3} }{2} }$

$\displaystyle\sf{ \implies \: \cos \theta = \cos \: \frac{\pi}{6} }$

$\displaystyle\sf{ \implies \: \theta = \: \frac{\pi}{6} \: \: and \: \frac{11\pi}{6} \: \: \big( \because \: \theta \: \in \: (0,2\pi) \big) }$

FINAL ANSWER

Hence the required solution is

$\boxed{ \: \: \displaystyle\sf{ \: \theta = \: \frac{\pi}{6} \: , \: \frac{5\pi}{6} \: , \: \frac{11\pi}{6} \: \: \big( \: where \: \theta \: \in \: (0,2\pi) \big) } \: \: }$

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Answer 2

Answer:

SOLUTION

TO DETERMINE

The solution set of the equation

4sin θ cos θ - 2cos θ - 2√3sin θ + √3 = 0

in the interval (0,2π)

EVALUATION

Here the given equation is

4sin θ cos θ - 2cos θ - 2√3sin θ + √3 = 0

We now solve for θ as below

4sin θ cos θ - 2cos θ - 2√3sin θ + √3 = 0

⇒ 2cos θ ( 2sin θ - 1 ) - √3 ( 2sin θ - 1 ) = 0

⇒ ( 2sin θ - 1 ) ( 2cos θ - √3 ) = 0

Now 2sin θ - 1 = 0 gives

Again ( 2cos θ - √3 ) = 0 gives

FINAL ANSWER

Hence the required solution is

━━━━━━━━━━━━━━━━

Learn more from Brainly :-

1. If cosθ+secθ=√2,find the value of cos²θ+sec²θ

brainly.in/question/25478419

2. Value of 3 + cot 80 cot 20/cot80+cot20 is equal to

brainly.in/question/17024513

3. In a triangle, prove that (b+c-a)(cotB/2+cotC/2)=2a×cotA/2

Step-by-step explanation: