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Question: How do you find the solution to \({{\tan }^{2}}\theta +4\tan \theta -12=0\) if \(0\le \theta \le 360...

How do you find the solution to tan⁡2θ+4tan⁡θ−12=0{{\tan }^{2}}\theta +4\tan \theta -12=0 if 0≤θ≤360?0\le \theta \le 360?

Explanation

Solution

We will add and subtract necessary terms to and from the given trigonometric equation. Then we will take the greatest common factors out. Then, we will find out the factors of the equation. From that, we will be able to find the solution.

Complete step by step solution:
Let us consider the given trigonometric identity tan⁡2θ+4tan⁡θ−12=0.{{\tan }^{2}}\theta +4\tan \theta -12=0.
We will add and subtract 2tan⁡θ2\tan \theta on the left-hand side of the given equation.
We will get tan⁡2θ+6tan⁡θ−2tan⁡θ−12=0.{{\tan }^{2}}\theta +6\tan \theta -2\tan \theta -12=0.
Let us take the common termtan⁡θ\tan \theta from the first term and the second term out and take the common term −2-2 from the third term and the fourth term.
Then, we will get the equation as tan⁡θ(tan⁡θ+6)−2(tan⁡θ+6)=0.\tan \theta \left( \tan \theta +6 \right)-2\left( \tan \theta +6 \right)=0.
As we can see, there is still a common term. The term is tan⁡θ+6.\tan \theta +6. Let us take this term out.
We will get the equation in the form of (tan⁡θ+6)(tan⁡θ−2)=0.\left( \tan \theta +6 \right)\left( \tan \theta -2 \right)=0.
So, from what we have learnt earlier, we can say tan⁡θ+6=0\tan \theta +6=0 or tan⁡θ−2=0.\tan \theta -2=0.
Let us consider the term tan⁡θ+6.\tan \theta +6. We will get tan⁡θ+6=0.\tan \theta +6=0.
From this equation, we will get tan⁡θ=−6.\tan \theta =-6.
And from this step, we can find the value of θ\theta for which tan⁡θ+6=0.\tan \theta +6=0.
We can find it as θ=tan⁡−1−6=−80.54∘=360−80.54=279.46∘.\theta ={{\tan }^{-1}}-6=-80.54{}^\circ =360-80.54=279.46{}^\circ .
Also, we will get θ=279.46−180=99.46∘.\theta =279.46-180=99.46{}^\circ .
This is because the Tangent function is negative in the second quadrant and the fourth quadrant.
Let us consider the other term tan⁡θ−2.\tan \theta -2. It will give us tan⁡θ−2=0.\tan \theta -2=0.
This will lead us to obtain tan⁡θ=2.\tan \theta =2.
Thus, we can find the value of θ\theta for which tan⁡θ−2=0\tan \theta -2=0 by θ=tan⁡−12=63.43∘.\theta ={{\tan }^{-1}}2=63.43{}^\circ .
Also, we will get θ=180+63.43=243.43∘.\theta =180+63.43=243.43{}^\circ .
This is because the Tangent function is positive in the first quadrant and the second quadrant.

Hence the solution of the given equation is 0≤θ≤360,θ=63.43∘,99.46∘,243.43∘,279.46∘.0\le \theta \le 360, \theta =63.43{}^\circ ,99.46{}^\circ ,243.43{}^\circ ,279.46{}^\circ .

Note: We can solve this equation by putting tan⁡θ=x.\tan \theta =x. The equation will become x2+4x−12=0.{{x}^{2}}+4x-12=0. This can be factorized to get (x−2)(x+6)=0.\left( x-2 \right)\left( x+6 \right)=0. So, we will get x=2x=2 or x=−6.x=-6. Therefore, tan⁡θ=2\tan \theta =2 or tan⁡θ=−6.\tan \theta =-6.