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Question: If \[7{\sin ^2}x + 3{\cos ^2}x = 4\], show that \(\tan x = \dfrac{1}{{\sqrt 3 }}\)....

If 7sin⁡2x+3cos⁡2x=47{\sin ^2}x + 3{\cos ^2}x = 4, show that tan⁡x=13\tan x = \dfrac{1}{{\sqrt 3 }}.

Explanation

Solution

We can split the left hand side of the given equation. 77 in the equation can be replaced as 4+34 + 3. So we can take three as common on the LHS. Then we can apply the trigonometric relation of sin⁡2x+cos⁡2x{\sin ^2}x + {\cos ^2}x. Now we can simplify the equation. Using the sine value we can find the angle and so we can find the tangent value.

Formula used: For angles in radians we have,
sin⁡π6=12\sin \dfrac{\pi }{6} = \dfrac{1}{2}
tan⁡π6=13\tan \dfrac{\pi }{6} = \dfrac{1}{{\sqrt 3 }}
For any angle xx we have, sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1

Complete step-by-step solution:
Given that 7sin⁡2x+3cos⁡2x=47{\sin ^2}x + 3{\cos ^2}x = 4
Since 7sin⁡2x=4sin⁡2x+3sin⁡2x7{\sin ^2}x = 4{\sin ^2}x + 3{\sin ^2}x
We can write,
\Rightarrow$$$4{\sin ^2}x + 3{\sin ^2}x + 3{\cos ^2}x = 4$$ Taking 3commonfromthesecondandthirdtermswehave,common from the second and third terms we have, \Rightarrow4{\sin ^2}x + 3({\sin ^2}x + {\cos ^2}x) = 4$$ We know $${\sin ^2}x + {\cos ^2}x = 1$$ Substituting this we have, $\Rightarrow4{\sin ^2}x + 3 \times 1 = 4 \Rightarrow 4{\sin ^2}x + 3 = 4 Subtracting $3$ from both sides we get, $\Rightarrow$$$4{\sin ^2}x + 3 - 3 = 4 - 3
⇒4sin⁡2x=1\Rightarrow 4{\sin ^2}x = 1
Dividing both sides by 44 we get,
\Rightarrow$$${\sin ^2}x = \dfrac{1}{4}$$ Taking square roots on both sides we have, \Rightarrow$$$\sin x = \pm \dfrac{1}{2}Consideringpositiverootwehave Considering positive root we have\sin x = \dfrac{1}{2}$$
This gives x=π6x = \dfrac{\pi }{6}, since sin⁡π6=12\sin \dfrac{\pi }{6} = \dfrac{1}{2}
Using this we can find the value of tan⁡x\tan x.
So tan⁡x=tan⁡π6=13\tan x = \tan \dfrac{\pi }{6} = \dfrac{1}{{\sqrt 3 }}
Hence we had proved the given statement.

Additional Information: In a right angled triangle with one of the non-right angles θ\theta , sin⁡θ\sin \theta refers to the ratio of the opposite side of the angle θ\theta to the hypotenuse and cos⁡θ\cos \theta refers to the ratio of the adjacent side of the angle θ\theta to the hypotenuse. Whereas the tan⁡θ\tan \theta refers to the ratio of the opposite side of the angle θ\theta to the adjacent side of the angle θ\theta .

Note: Actually sin function is a periodic function with period 2π2\pi . But here we considered only one angle of sine which gives the value 12\dfrac{1}{2}. So we got the value π6\dfrac{\pi }{6} for xx.