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Question

Question: Solve \[{\tan ^2}3\theta = 3\]...

Solve tan23θ=3{\tan ^2}3\theta = 3

Explanation

Solution

Here we are asked to solve the given equation which contains trigonometric functions. Using the basic definition of trigonometric conversions to convert to the simplest forms possible. Preferably convert them into terms of single or double angles so that the computation is feasible for us to handle. After these conversions we can be able to simplify the equation easily. Keep in mind that this is not a polynomial equation to solve for any unknown variable; it contains a trigonometric function thus we have to solve it for the value of theta.

Complete answer:
It is given that tan23θ=3{\tan ^2}3\theta = 3 We aim to solve this equation for the value of theta. Let us first expand the given function using trigonometric identities.
We know tan2θ=sec2θ1{\tan ^2}\theta = {\sec ^2}\theta - 1
So,
sec23θ13=0{\sec ^2}3\theta - 1 - 3 = 0
sec23θ=4\Rightarrow {\sec ^2}3\theta = 4

Also,
sec2θ=1cos2θ{\sec ^2}\theta = \dfrac{1}{{{{\cos }^2}\theta }}
Then,
1cos23θ=4\dfrac{1}{{{{\cos }^2}3\theta }} = 4
cos23θ=14\Rightarrow {\cos ^2}3\theta = \dfrac{1}{4}
cos23θ=12.2\Rightarrow {\cos ^2}3\theta = \dfrac{1}{{2.2}}
2cos23θ=12\Rightarrow 2{\cos ^2}3\theta = \dfrac{1}{2}
From known properties, 2cos2θ=1+cos2θ2{\cos ^2}\theta = 1 + \cos 2\theta
Then,
1+cos6θ=121 + \cos 6\theta = \dfrac{1}{2}
cos6θ=12\Rightarrow \cos 6\theta = - \dfrac{1}{2}
Using the angle values from the table we get,

cos0\cos {0^ \circ }cos30\cos {30^ \circ }cos45\cos {45^ \circ }cos60\cos {60^ \circ }cos90\cos {90^ \circ }
1132\dfrac{{\sqrt 3 }}{2}12\dfrac{1}{{\sqrt 2 }}12\dfrac{1}{2}00

So,
cos6θ=cosπ3\Rightarrow \cos 6\theta = - \cos \dfrac{\pi }{3}
Now again we write it as,
cos6θ=cos(ππ3)\Rightarrow \cos 6\theta = \cos (\pi - \dfrac{\pi }{3})
cos6θ=cos(2π3)\Rightarrow \cos 6\theta = \cos (\dfrac{{2\pi }}{3})
Now we know that if cosx=cosy\cos x = \cos y then x=2nπ±yx = 2n\pi \pm y wheren=0,±1,±2n = 0, \pm 1, \pm 2
We get,
6θ=2nπ±2π36\theta = 2n\pi \pm \dfrac{{2\pi }}{3}
θ=nπ3±π9\theta = \dfrac{{n\pi }}{3} \pm \dfrac{\pi }{9}
Therefore, θ=π3(n±13)\theta = \dfrac{\pi }{3}(n \pm \dfrac{1}{3}) where n=0,±1,±2n = 0, \pm 1, \pm 2

Note:
To solve these types of problem we must need to know all the basic trigonometric identities because trigonometric identities are the key to solve this problem. It plays a major role in simplifying or breaking the given equation into a simple form which makes the solving easier. We must give more attention to using the correct trigonometric identity because some identities may make the simpler form into a complex form. Another important thing that we need to know is the trigonometric ratio table here we are solving the given function for the value of theta it can only be found by using the trigonometric ratio table.