3cos2- -2-3sin-cos- -3sin2- -0-

3cos ^2θ 2√(3)sinθcosθ 3sin^2=0 #160;Dividing by cos ^2θ #160; 3 2√(3)tanθ 3 tan^2θ=0 #160; 3tan^2θ+2√(3)tanθ 3=0 #160; 3tan ...

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Byju's Answer

Standard XII

Mathematics

Principal Solution of Trigonometric Equation

3cos2θ -2√3si...

Question

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

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Solution

$3 {cos}^{2} θ - 2 \sqrt{3} sin θ cos θ - 3 {sin}^{2} = 0$
Dividing by ${cos}^{2} θ$
$3 - 2 \sqrt{3} tan θ - 3 {tan}^{2} θ = 0$
$3 {tan}^{2} θ + 2 \sqrt{3} tan θ - 3 = 0$
$3 {tan}^{2} θ + 3 \sqrt{3} tan θ - \sqrt{3} tan θ - 3 = 0$
$3 tan θ (tan θ + \sqrt{3}) - \sqrt{3} (tan θ + \sqrt{3}) = 0$
$(3 tan θ - \sqrt{3}) (tan θ + \sqrt{3}) = 0$

$tan θ = \frac{1}{\sqrt{3}}$ or $tan θ = - \sqrt{3}$

$θ = n π + \frac{π}{6}$ or $θ = n π - \frac{π}{3}$

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Similar questions

Q. Solve

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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3cos2θ−2√3sinθcosθ−3sin2θ=0.

3cos2θ−2√3sinθcosθ−3sin2=0 Dividing by cos2θ 3−2√3tanθ−3tan2θ=0 3tan2θ+2√3tanθ−3=0 3tan2θ+3√3tanθ−√3tanθ−3=0 3tanθ(tanθ+√3)−√3(tanθ+√3)=0(3tanθ−√3)(tanθ+√3)=0 tanθ=1√3 or tanθ=−√3 θ=nπ+π6 or θ=nπ−π3

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