How do you find the general solutions for #cos^2(x) + Sin(x)=0#?

To solve this type of equation you need to start memorising trig identities.

Given equation: #cos^2(x) + sin(x) = 0#..........(1)

If we can change the cos into sine then we have a quadratic equation which is very solvable. Once #sin(x)# is found then by substitution in (1) we can find #cos(x)#.

Using the Trig identity:
#sin^2(x)+cos^2(x) = 1# ..................(2)
#" "#(Note that #" "sin^2(x)" "# is #" "[sin(x)]^2#)

Rewrite (2) as: #cos^2(x)=1 - sin^2(x)# ............(3)

Substitute (3) into (1) giving:

#{-sin^2(x)} + sin(x) + 1 =0# ........................(4)

To reduce confusion: let #sin(x)# be represented by#" " u#

so we have #(-u)^2+u+1=0#

Using standard quadratic form of
#ax^2 +bx+c=0 ->" "x= (-b+-sqrt(b^2-4ac))/(2a)# giving:

#a=(-1)," " b=1," ",c=1#

#u=(-1+-sqrt((1)^2-(4)(-1)(1)))/(2(-1)) #

so #u=sin(x)= (-1 +-sqrt(5))/((-2) #.............(5)

Substitute (5) in (1) to find #cos(x)#

Once you have the final answer do not forget that this cyclic. So it repeats every turn of a circle which is #2pi# radians. So your answer would be:

#n(2pi) times "answer"# were n is a counting number and #n>=0#
that is #" "##n in N#. Some people state that zero does not belong to the counting numbers.