Trignometric identities question help

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BabzY
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For solutions in the range -180 < theta < 180

The answer is 30 and 150 degrees but i would like to see the method for this question, any help is appreciated
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RDKGames
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(Original post by BabzY)
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For solutions in the range -180 < theta < 180

The answer is 30 and 150 degrees but i would like to see the method for this question, any help is appreciated
Do you know how to firstly solve this equation in terms of a single trigonometric function?? I.e. what identity can you use to reduce this equations into a quadratic in terms of a single trig function?
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the bear
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there is a well known relationship between tan2 and sec2
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BabzY
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1 + tan^2 theta = sec^2 theta , using that i get a quadratic equation by subbing sec^2 theta -1 for tan^2 theta however when i work it out my answer doesn't match the one in the book
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RDKGames
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(Original post by BabzY)
1 + tan^2 theta = sec^2 theta , using that i get a quadratic equation by subbing sec^2 theta -1 for tan^2 theta however when i work it out my answer doesn't match the one in the book
Okay, so what do you do when you get \sec^2(\theta)-5\sec(\theta)+2=0 ??
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the bear
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is your calculator set to degrees ?

:holmes:
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jackdavidson1874
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(Original post by RDKGames)
Okay, so what do you do when you get \sec^2(\theta)-5\sec(\theta)+2=0 ??
You would factorise it as a quadratic and then set both brackets equal to 0 and solve
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BabzY
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hahaha, thanks guys. Was using the correct method but forgot all about the calculator settings
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