Solve 3cos 2- -2-3sin-cos- -3sin 2- -0

Given, 3cos ^ 2 θ #160; √( 3 )sinθ #160;cosθ #160; 3sin ^ 2 θ = 0 ⇒ 3cos 2θ #160; √( 3 )sin 2θ #160; =0 ⇒√( 3 )cos 2θ #160; sin 2θ ...

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Standard XII

Mathematics

Property 1

Solve 3cos ...

Question

Solve $3 {cos}^{2} θ - 2 \sqrt{3} sin θ cos θ - 3 {sin}^{2} θ = 0$

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Solution

Given, $3 {cos}^{2} θ - \sqrt{3} sin θ cos θ - 3 {sin}^{2} θ = 0 \Rightarrow 3 cos 2 θ - \sqrt{3} sin 2 θ = 0 \Rightarrow \sqrt{3} cos 2 θ - sin 2 θ = 0$
$\Rightarrow \frac{\sqrt{3}}{2} cos 2 θ - \frac{1}{2} sin 2 θ = 0$
$\Rightarrow sin (\frac{π}{3} - 2 θ) = 0$
$\Rightarrow \frac{π}{3} - 2 θ = n π \Rightarrow 2 θ = n π - \frac{π}{3}$

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Similar questions

3 {cos}^{2} θ - 2 \sqrt{3} sin θ cos θ - 3 {sin}^{2} θ = 0

Q. If

\sqrt{3} {cos}^{2} θ - 2 \sqrt{3} sin θ cos θ - 3 {sin}^{2} θ = 0

, then

$Solve the following equations : (i) s i n^{2} θ - c o s θ = \frac{1}{4} (i i) 2 c o s^{2} θ - 5 c o s θ + 2 = 0 (i i i) 2 s i n^{2} x + \sqrt{3} c o s x + 1 = 0 (i v) 4 s i n^{2} θ - 8 c o s θ + 1 = 0 (v) t a n^{2} x + (1 - \sqrt{3}) t a n x - \sqrt{3} = 0 (v i) 3 c o s^{2} θ - 2 \sqrt{3} s i n θ c o s θ - 3 s i n^{2} θ = 0 (v i i) c o s 4 θ = c o s 2 θ$

Q.
In equation

3^{s i n^{2} θ} + 3^{c o s^{2} θ} \geq 2 \sqrt{3}

is true

Q. Arrange the following equations in decreasing order of their number of solutions in

[0, 2 π]

3 {sin}^{2} θ + 4 {cos}^{2} θ = 5

II)

4 {sin}^{2} θ + 3 {cos}^{2} θ = \frac{7}{2}

III)

3 {sin}^{2} θ + 4 {cos}^{2} θ = 4

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Solve 3cos2θ−2√3sinθcosθ−3sin2θ=0

Given, 3cos2θ−√3sinθcosθ−3sin2θ=0⇒3cos2θ−√3sin2θ=0⇒√3cos2θ−sin2θ=0⇒√32cos2θ−12sin2θ=0⇒sin(π3−2θ)=0⇒π3−2θ=nπ⇒2θ=nπ−π3

Solve $3 {cos}^{2} θ - 2 \sqrt{3} sin θ cos θ - 3 {sin}^{2} θ = 0$