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jek_recluse [69]
3 years ago
6

Use a graphing utility to solve the equation on the interval 0°< x < 360°. Express the solution(s) rounded to one decimal

place.
cos^2 x + cos x - 1 = 0
Mathematics
1 answer:
Ludmilka [50]3 years ago
8 0
<span>cos^2 x + cos x - 1 = 0, and if X=cosx, x=arccosX
let's find X
X²+X-1=0, delta=1-4(-1)=5, X= -1-sqrt5 /2 or </span><span> X= -1+ sqrt5 /2, but </span><span>0°< x < 360°, 
so we must take </span>X= -1+ sqrt5 /2, and <span> x=arccosX = 51.82°</span>
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