<p>Can anyone help me out w/ these? They're relatively simple, but I've forgotten a lot over the summer and just want to make sure I didn't leave out answers or made careless mistakes. </p>
<p>"Solve each of the equations for 0 lessthan= x lessthan= 2pi. Remember t odouble the domain when solving for a double angle." </p>
<p>4cos^(2)x-3=0</p>
<p>sin2x= -√3/2 </p>
<p>Lastly, "Find the remainders on division of x^5-x^4+x^3+2x^2-x+4 by x^3+1</p>
<p>Thanks!!</p>
<p>Dude. Basic trig. For the first. Send the 3 to the other side and divide by 4. Now take the square root and solve. Please tell me u can solve from here. </p>
<p>For the second. You have to options. Apply half angle formula, simplify, and solve
Or substitute 2x=y. Solve for Y. Put the values of Y in 2x=y. Ie divide the values of y by 2 in order to get the value of x.
Remember to discard anything outside of domain.</p>
<p>PS. U can directly solve for x. By finding 2x as is then dividing by two. But i find it easier to subsitute for a little while</p>
<p>Sent from GT-I9300 using CC</p>
<p>@Gabiii:
For the last one, use polynomial long division. (You can’t use synthetic division here. I assume you know why.)</p>
<p>We have:
4cos^(2)x-3=0 0<x<2pi
Solution:
cos^(2)x=3/4
cos(x)= √3/2 cos(x)= -√3/2
x=arcos(√3/2) + 2pi*n n=0,1,…
x=pi/6</p>
<p>cos(x)= -√3/2
x=arcos(-√3/2) + 2pi<em>n n=0,1,…
x=pi- arcos(√3/2) + 2pi</em>n n=0,1,…
x=5pi/6 </p>
<p>We have:
sin2x= -√3/2
Solution:
2x=arcsin(-√3/2) + 2pi<em>n
2x= - pi/3 + 2pi</em>n n=0,1,…
x= -pi/6 + pi*n n=0,1,…
n=0 x=-pi/6
n=1 x= 5pi/6</p>
<p>We have:
Lastly, "Find the remainders on division of x^5-x^4+x^3+2x^2-x+4 by x^3+1</p>
<p>Solution:
x^5-x^4+x^3+2x^2-x+4|_____x^3+1</p>
<h2>x^5 + x^2 x^2- x + 1</h2>
<p>-x^4 + x^3 +x^2-x+4</p>
<h2>- x^4 -x</h2>
<pre><code> x^3 +x^2 +4
x^3 + +1
</code></pre>
<hr>
<pre><code> x^2 +3
</code></pre>
<p>So the remainders on division of x^5-x^4+x^3+2x^2-x+4 by x^3+1 is x^2 +3</p>