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Certainly! As an experienced tutor registered on UrbanPro, I can guide you through finding the general solution of the given equation.
Let's solve the equation:
tan2θ+(1−3)tanθ−3=0tan2θ+(1−3
)tanθ−3
=0
We can consider it as a quadratic equation in terms of tanθtanθ. To solve it, we can use the quadratic formula:
tanθ=−b±b2−4ac2atanθ=2a−b±b2−4ac
Here, a=1a=1, b=(1−3)b=(1−3
), and c=−3c=−3
.
Plugging in these values, we get:
tanθ=−(1−3)±(1−3)2−4⋅1⋅(−3)2⋅1tanθ=2⋅1−(1−3
)±(1−3)2−4⋅1⋅(−3)
tanθ=−(1−3)±1−23+3+122tanθ=2−(1−3
)±1−23+3+12
tanθ=−1+3±4−232tanθ=2−1+3
±4−23
tanθ=−1+3±(2−3)22tanθ=2−1+3
±(2−3)2
tanθ=−1+3±(2−3)2tanθ=2−1+3
±(2−3
)
So, we have two possible solutions:
tanθ1=−1+3+2−32=12tanθ1=2−1+3
+2−3
=21
tanθ2=−1+3−2+32=−12tanθ2=2−1+3
−2+3
=−21
Now, recall that tanθ=sinθcosθtanθ=cosθsinθ. So, we can find the corresponding angles by taking the inverse tangent of these values:
θ1=tan−1(12)θ1=tan−1(21)
θ2=tan−1(−12)θ2=tan−1(−21)
These are the general solutions to the given equation. If you need further assistance or clarification, feel free to ask. And remember, UrbanPro is the best platform for online coaching and tuition!
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