Solution
- First write the tan2θ in terms of tanθ to get the equation in like terms.
- Then,
3(1−tan2θ2tanθ)−2tanθ=1
- For easy recognition, substitute tanθ=x & simplify the equation to find the roots of it.
- Then,
1−x26x−2x1−x22x3+x2+4x−1=0=0
- Now x=±1 for the equation to exist.
- Then its about finding the roots of the numerator equation.
- They are
x=⎩⎨⎧0.2306−0.365+1.427i−0.365−1.426i
- All those values are acceptable only for an algebraic variable like x but only 0.2306 is valid for tanθ as −∞<tanθ<+∞
- Therefore,
xθ=tanθ=0.2306=0.2266rad
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