solve the following equation for x, where 0
b) 3 sec²x - 4 = 0
So I know this is what we have to do using the trig identities:
3 sec²x - 4 = 0
3 (1 + tan²x) - 4 = 0
3 + 3tan²x - 4 = 0
3tan²x - 1 =0
Then I get stuck... because I can't expand this to solve for x...
someone please help me? :(
So I know this is what we have to do using the trig identities:
3 sec²x - 4 = 0
3 (1 + tan²x) - 4 = 0
3 + 3tan²x - 4 = 0
3tan²x - 1 =0
Then I get stuck... because I can't expand this to solve for x...
someone please help me? :(
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You don't even need that identity. Isolate the sec^2x
3sec²x = 4
sec²x = 4/3
Take the square root
secx = √(4/3)
secx = 2/√(3)
secx = 1/cosx
1/cosx = 2/√(3)
cosx = √(3)/2
Now solve for x using inverse trig functions
arccos(√(3)/2) = x
x = pi/6, 11pi/6
3sec²x = 4
sec²x = 4/3
Take the square root
secx = √(4/3)
secx = 2/√(3)
secx = 1/cosx
1/cosx = 2/√(3)
cosx = √(3)/2
Now solve for x using inverse trig functions
arccos(√(3)/2) = x
x = pi/6, 11pi/6