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Question: Solve the following equations: \(1 + {\sin ^2}\theta = 3\sin \theta .\cos \theta \)...

Solve the following equations:
1+sin⁡2θ=3sin⁡θ.cos⁡θ1 + {\sin ^2}\theta = 3\sin \theta .\cos \theta

Explanation

Solution

Hint: To solve this question we have to start from given but we are not able to how to proceed further so in place of 1 we will use (∵sin⁡2θ+cos⁡2θ=1)\left( {\because {{\sin }^2}\theta + {{\cos }^2}\theta = 1} \right) then simple mathematical calculation further to get answer.

Complete step-by-step answer:
We have given
1+sin⁡2θ=3sin⁡θcos⁡θ1 + {\sin ^2}\theta = 3\sin \theta \cos \theta
⇒1+sin⁡2θ−3sin⁡θ.cos⁡θ=0\Rightarrow 1 + {\sin ^2}\theta - 3\sin \theta .\cos \theta = 0
From here we can’t find any way of proceeding further so use standard results to proceed.
⇒sin⁡2θ+cos⁡2θ+sin⁡2θ−3sin⁡θ.cos⁡θ=0\Rightarrow {\sin ^2}\theta + {\cos ^2}\theta + {\sin ^2}\theta - 3\sin \theta .\cos \theta = 0 [∵sin⁡2θ+cos⁡2θ=1]\left[ {\because {{\sin }^2}\theta + {{\cos }^2}\theta = 1} \right]
⇒2sin⁡2θ−3sin⁡θ.sin⁡θ+cos⁡2θ=0\Rightarrow 2{\sin ^2}\theta - 3\sin \theta .\sin \theta + {\cos ^2}\theta = 0
Now this is quadratic in trigonometric function so we will now factorize it .
⇒2sin⁡2θ−2sin⁡θ.cos⁡θ−sin⁡θ.cos⁡θ+cos⁡2θ=0\Rightarrow 2{\sin ^2}\theta - 2\sin \theta .\cos \theta - \sin \theta .\cos \theta + {\cos ^2}\theta = 0
⇒2sin⁡θ(sin⁡θ−cos⁡θ)−cos⁡θ(sin⁡θ−cos⁡θ)=0\Rightarrow 2\sin \theta \left( {\sin \theta - \cos \theta } \right) - \cos \theta \left( {\sin \theta - \cos \theta } \right) = 0
On taking (sin⁡θ−cos⁡θ)\left( {\sin \theta - \cos \theta } \right) common we get
⇒(sin⁡θ−cos⁡θ)(2sin⁡θ−cos⁡θ)=0\Rightarrow \left( {\sin \theta - \cos \theta } \right)\left( {2\sin \theta - \cos \theta } \right) = 0
Therefore,
(sin⁡θ−cos⁡θ)=0……(i)\left( {\sin \theta - \cos \theta } \right) = 0 \ldots \ldots \left( {\text{i}} \right) or,
(2sin⁡θ−cos⁡θ)=0\left( {2\sin \theta - \cos \theta } \right) = 0 ……(ii) \ldots \ldots \left( {{\text{ii}}} \right)
From equation (i)
sin⁡θ=cos⁡θ tan⁡θ=1[∵tan⁡π4=1] ∴θ=π4  \sin \theta = \cos \theta \\\ \tan \theta = 1 \left[ {\because \tan \dfrac{\pi }{4} = 1} \right] \\\ \therefore \theta = \dfrac{\pi }{4} \\\
From equation (ii)
2sin⁡θ−cos⁡θ=0 tan⁡θ=12 θ=tan⁡−1(12)  2\sin \theta - \cos \theta = 0 \\\ \tan \theta = \dfrac{1}{2} \\\ \theta = {\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right) \\\

Note: Whenever you get this type of question where we can’t find any way of proceeding then we have used our brain wisely as written in hint to proceed in this question. We can find the solution of the quadratic equation by other methods as well but those methods make calculation difficult. This type of question pushes us to think and make us prepare for advanced level preparation.