Trignometric identities question help

#1

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
0
5 years ago
#2
(Original post by BabzY)

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?
0
5 years ago
#3
there is a well known relationship between tan2 and sec2
0
#4
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
0
5 years ago
#5
(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 ??
0
5 years ago
#6
is your calculator set to degrees ?

0
5 years ago
#7
(Original post by RDKGames)
Okay, so what do you do when you get ??
You would factorise it as a quadratic and then set both brackets equal to 0 and solve
0
#8
hahaha, thanks guys. Was using the correct method but forgot all about the calculator settings
0
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