Question

if 7 sin² thita +3 cos² thita=4, find the value of thita​

Answer 1

7 sin² ∅ + 3 cos² ∅ = 4 .

⇒ 4 sin²∅ + 3 sin²∅ + 3 cos²∅ = 4 .

⇒ 4 sin²∅ + 3( sin²∅ + cos²∅ ) = 4 .

⇒ 4 sin²∅ + 3( 1 ) = 4 . [ ∵ sin²∅ + cos²∅ = 1 ] .

⇒ 4 sin²∅ + 3 = 4 .

⇒ 4 sin²∅ = 4 - 3 .

⇒ 4 sin²∅ = 1 .

⇒ sin²∅ = 1/4 .

⇒ sin ∅ = √(1/4) .

∴ sin ∅ = 1/2 .

But, sin 30° = 1/2 .

Then, sin ∅ = sin 30° .

Then, tan 30° = 1/√3 .

Hence, it is proved .

7sin²θ + 3cos²θ = 4

7sin²θ + ౩(1-sin²θ) = 4

7sin²θ+3-3sin²θ = 4

4sin²θ+3 = 4

4sin²θ=1

sin²θ = 1/4

sinθ = +1/2

sinθ = + π/6

Therefore tan θ = + π/6

then we get tan θ = + 1/√3

As the required answer is tan θ = + 1/√3 (here we considered θ =+ π/6)

Answer 2

Given :

$\\ \large\red\bigstar\bf{ 7sin^2 \theta +3 cos^2 \theta =4}$

To find :

$\\ \large\red\bigstar\bf{ the \: value \: of \: \theta =?}$

Solution :

$\implies \large \bf \: 7 {sin }^{2} \theta \: + 3 {cos}^{2} \theta = 4 \\ \\ \implies\bf \red{ \underline{by \: using \: identity \: {sin}^{2} \theta + {cos}^{2} \theta = 1}} \\ \\ \implies \large \bf \: 7(1 - {cos}^{2} \theta) + 3 {cos}^{2} \theta = 4 \\ \\ \implies \large \bf \: 7 - 7 {cos}^{2} \theta + 3 {cos}^{2} \theta = 4 \\ \\ \implies \large \bf \: 7 - 4 {cos}^{2} \theta = 4 \\ \\ \implies \large \bf \: 4 {cos}^{2} \theta = 3 \\ \\ \implies \large \bf \: {cos}^{2} \theta = \bigg( { \frac{ \sqrt{3} }{2} } \bigg)^{2} \\ \\ \\ \therefore \large \sf \: the \: general \: solution \: of \theta \: is \boxed{ \sf{ \: n\pi \pm \frac{\pi}{6} }}$