Trigonometry!
Can someone help me? I can't remember how to handle the part in the brackets here:
Sec(3x+ Pi/6 )= -2/(3^1/2)
Sec(3x+ Pi/6 )= -2/(3^1/2)
Original post by Carokelly123
Can someone help me? I can't remember how to handle the part in the brackets here:
Sec(3x+ Pi/6 )= -2/(3^1/2)
Sec(3x+ Pi/6 )= -2/(3^1/2)
1) Take your equation and write it in terms of cosine
2) Whatever your range for x is, find the corresponding range for 3x + pi/6
3) Find the values that satisfy 3x+pi/6 for the range
4) Find x
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