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PDL without converse operator -

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Clash Royale CLAN TAG #URR8PPP up vote 1 down vote favorite I found this exercise (5.14) in the book "Dynamic Logic" by Harel, Kozen and Tiuryn. I have no clue how to construct the mentioned model in the exercise body. Construct the Kripke model, such that the operator $<alpha>$ is not continous. In PDL with - (converse operator), the map $phi rightarrow <alpha>phi$ is continuous with respect to the order of logical implication. That is, if $K$ is a Kripke frame, $A$ a (finite or infinite) set of formulas, and $phi$ a formula such that $m_K(phi) = sup_psi in Am_K(psi)$ then $sup_psi in Am_K(<alpha>psi)$ exists and is equal to $m_K(<alpha>phi)$. We define $m_K(psi)$ as a set of states in $K$ satisfying formula $psi$. If program $alpha$ maps state $s_1$ to state $s_2$ then program $alpha-$ ($alpha$ with converse operator) maps $s_2$ to $s_1$. logic graph-theory modal-logic share | cite | improve this question edited Aug 26 at 17:2

Sum of roots of equation $x^4 - 2x^2 sin^2(displaystyle pi x/2) +1 =0$ is

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Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 My try: $$x^4-2x^2sin^2(fracpi x2)+1=0\x^4+1=2x^2left (1-cos^2left(fracpi x2right)right )\(x^2-1)^2=-2x^2cos^2left(fracpi x2right)\(x^2-1)^2+2x^2cos^2left(fracpi x2right)=0\x^2-1=0,textand, 2x^2cos^2left(fracpi x2right)=0$$ I am stuck , I am confused now what to do now algebra-precalculus trigonometry roots share | cite | improve this question edited Aug 26 at 15:11 TheSimpliFire 10.7k 6 20 54 asked Aug 26 at 14:08 user580093 64 1 5 Please format your equations using Mathjax as this makes it a lot easier for us to read. math.meta.stackexchange.com/questions/5020/… – Jam Aug 26 at 14:11 3 You've actually done the hard part of the problem :) – TheSimpliFire Aug 26 at 14:15 Clearly the only real solutions are 1 and -1. But what about complex solutions? – dm63 Aug 26 at 15:01 indeed, @Batominovski, the roots must