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Question Number 57915 by Tinkutara last updated on 14/Apr/19

Answered by mr W last updated on 14/Apr/19

$$(1-{\tan }^2\theta )(1+{\tan }^2\theta )+2^{{\tan }^2\theta }=0$$$$lett={\tan }^2\theta ⩾0$$$$\Rightarrow 1-t^2+2^t=0$$$$\Rightarrow 2^t=t^2-1$$$$\Rightarrow t=3and3.4075$$$$\Rightarrow \theta =\pm {\tan }^{-1}\sqrt{3}=\pm \frac{\pi }{3}=\pm 60°$$$$\Rightarrow \theta =\pm {\tan }^{-1}\sqrt{3.4075}=\pm 61.55°$$

Commented by Tinkutara last updated on 14/Apr/19

t=3 can only be found with calculators?

Commented by mr W last updated on 14/Apr/19

$$t=3canbefoundbytry.othersolutions$$$$canonlybefoundthroughgraphic$$$$tools.$$

Commented by Tinkutara last updated on 15/Apr/19

Thanks Sir!

Answered by peter frank last updated on 14/Apr/19

$$a={tan}^2\theta$$$$(1-{\tan }^4\theta )+2^{\tan {}^2\theta }=0$$$$1-a^2+2^a=0$$$$usegraph$$$$a=-1.198,3,3.407$$$$\theta =\pm \frac{\pi }{3}$$$$\theta =\pm {\tan }^{-1}3.407$$$$$$

Commented by peter frank last updated on 14/Apr/19

Commented by mr W last updated on 14/Apr/19

$$sir,$$$$not\theta =\pm {\tan }^{-1}3.407$$$$but\theta =\pm {\tan }^{-1}\sqrt{3.407}$$

Commented by peter frank last updated on 14/Apr/19

$$thankyou$$

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