<p>1) Prove: (?1-sin x ÷ ?1+sin x) = lcos xl ÷ ( 1 + sin y)
The l l symbolize absolute value.... The ? marks are Radicals/Squareroot signs sorry!)</p>
<p>Solve the equations below over the Interval [0, 2n(pie))
2) cos2x + 5cosx= 2</p>
<p>3) tan?sin^2(Squared)? = tan?</p>
<p>Thanks Guys I guess these are kinda hard and I think the 2nd one is No solution but im not shure So if you could solve both of these that would be sweet</p>
<p>I don't know what the question marks mean, but the solution to the second one is:</p>
<p>cos(2x)=2cos^2(x)-1
substitute that in
then solve the resulting quadratic equation (with cos(x) as the variable)</p>
<p>thanks huggy I fixed the question marks! ANyone else kno</p>
<p>heyy wow can no one do this? If you can please do if you have the time ;)</p>
<p>first one you take the congugate of to figure it out.</p>
<p>[(1-sinx)^1/2(1+sinx)^1/2]/[(1+sinx)^1/2(1+sinx)^1/2]
[(1-sinx^2)^1/2]/(1+sinx)
[((cosx)^2)^1/2]/(1+sinx)
|cosx|/1+sinx</p>
<p>For the third one I have no clue what the equation is supposed to be saying, clear it up and I may attempt it.</p>
<p>Thanks man Umm the third questions is
tan(Theta)sin^2(theta)=tan(theta)</p>
<p>the ^2 is squared... Thanks Man!</p>
<p>3rd is when sinx = 0. where sinx = 1 would work except for the fact that when sinx=1 tanx=undefined. So that makes the roots 0, pi, and 2pi in the interval [0,2pi]</p>
<p>Reason being is that 0(n)=0. tanx = sinx/cosx. Wherever sinx=0 cosx is a defined number. So all we need to find is the points at which sinx=0</p>
<p>It can also be solved by moving the equation around but it leads to the same result.</p>
<p>tanxsinx^2-tanx = 0
tanx(sinx^2-1) = 0
-tanxcosx^2 = 0
-sinxcosx = 0
places where sinx = 0 because the places where cosx = 0 makes tanx undefined.</p>
<p>Hey Ice, The Intereval given is (2,n(Pie))</p>
<p>It doesn't matter the roots will still be 0, pi , and 2pi, but they will keep repeating around the circle.</p>
<p>so 0, pi +2npi, 2pi + 2npi where n is the number of complete rotations around the circle (If I understand your notation correctly)</p>