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cosx + sinxtanx = 2 find all solutions?
find all solutions of the equation in the interval 0,2pi
I understand the subject matter I just need help with understanding how to do this one equation
6 Answers
- 1 decade agoFavorite Answer
Change tan x into sinx / cosx. Then we have
cos x + sin x * (sin x / cos x) = 2
cos x + sin^2 x / cosx = 2
Multiply the cosx by (cosx / cosx) and combine the two fractions.
cos^2 x / cosx + sin^2 x / cosx = 2
(cos^2 x + sin^2 x) / cosx = 2
Remember: for every real number x we have sin^2 x + cos^2 x = 1.
(cos^2 x + sin^2 x) / cosx = 2 now becomes
1 / cos x = 2
Multply both sides by cos x
1 = 2cosx
Divide both sides by 2
cos x = 1/2
x = pi/3 or 5pi/3
- GilEspi.Lv 61 decade ago
cosx + sinx * tanx = 2
cosx + sinx * sinx/cosx = 2
cosx + sin^2x/cosx = 2
(cosx^2 + sin^2x) / cosx = 2
1/cosx = 2
1 = 2cosx
1/2 = cosx
arccos 1/2 = arccos cosx
x = pi/6 or 5pi/3
- Anonymous5 years ago
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cosx+sinxtanx=2 <=> (cos^2x + sin^2x)/cosx = 2 <=> 1/cosx = 2 <=> cosx = 1/2 = cosTT/3 =>x= TT/3
- 5 years ago
Where does the cosine come from in the transition from
cosx+sin^2x/cosx=2
to
(cos^2x+sin^2x)/cosx=2
?
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- Anonymous5 years ago
There are endless answers, for, as the graphs of all of these show, sec, cos, sin, tan, cos, csc, etc all go on forever. Just say there are endless answers.
- 1 decade ago
cos(x) + sin(x) * tan(x) = 2
cos(x) + sin(x) * sin(x) / cos(x) = 2
(cos(x)^2 + sin(x)^2) / cos(x) = 2
1 / cos(x) = 2
cos(x) = 1/2
x = pi/3 , 5pi/3
