Question

4cos5theta-2sqrt(3)=0

Answer 1

\[4\cos(5 \theta)-2 \sqrt{3}=0\]

Answer 2

\[2\cos(5 \theta)= \sqrt{3}\]

Answer 3

\[\cos(5 \theta) = \frac{ \sqrt{3} }{ 2 }\]

Answer 4

\[5 \theta = \cos^{-1}(\frac{ \sqrt{3} }{ 2 })\]

Answer 5

\[\theta = \frac{ 1 }{ 5 }\cos^{-1}(\frac{ \sqrt{3} }{ 2 })\]

Answer 6

ok I need to give all solutions

Answer 7

arccos(sqrt3/2)=30 degree or pi/6

Answer 8

then multiply by 1/5

Answer 9

well as you can see that you have a positive angle, which means you have to calculate the angles in the first and fourth quarter, and add 5x theta round

Answer 10

alright thanks! I got it. one quick question: would the problem 4cos^2=3=0 change any since there is no theta?

Answer 11

yes the answer would be different of you square the cosine

Answer 12

there would also be more solutions correct?

Answer 13

if*

Answer 14

yes

Answer 15

ok and for sin 1/3t=1/2 would one solution be pi/2?

Answer 16

sorry back

Answer 17

\[\sin(\frac{ 1 }{ 3t })=\frac{ 1 }{ 2 }\]

Answer 18

\[\frac{ 1 }{ 3t } = \sin^{-1}(0.5)\]

Answer 19

\[t = \frac{ 3 }{ \arcsin(0.5) }\]

Answer 20

\[t = 3arccsc(0.5)\]

Answer 21

i think one of the purposes of this question is to teach you the difference between inverse of a function and reciprocal