Question 870286
I am learning something important from this problem (and from the wiki I found once I figured out how to ask the question).
Two of your problems (aii and bii) were easy to answer without much algebra and/or trigonometry.
The other two were clearly related, but required more algebra work.
There had to be a common strategy to solve all four problems,
and using trigonometric identities had to be the key,
but I could not see how to use trigonometric identities to get where I wanted to get.
A quick internet search led me to the common strategy to solve the problems.
 
EACH PROBLEM AS A SEPARATE STRUGGLE:
(a)(ii) 
{{{2sin(x)+cos(x)=0}}}<--->{{{2sin(x)=-cos(x)}}}<--->{{{sin(x)/cos(x)=-1/2}}}<--->{{{tan(x)=-1/2}}}
From there we know that {{{x=-0.43648}}} (in radians, or {{{-26.5651^o}}} in degrees) is an answer.
Of course we know that there are infinite answers, {{{pi}}} (in radians) apart from each other,
because {{{y=tan(x)}}} is a periodic function with a period of {{{pi}}} (with {{{x}}} expressed in radians).

We could say that all of our approximate answers can be expressed as
{{{highlight(x=k*pi-0.43648)}}} (with {{{highlight(K=integer)}}} and {{{x}}} measured in radians).
The same strategy can be used to solve (b)(ii).
(I found {{{highlight(x=2.6779+k*pi)}}} as a general solution).
 
(a)(i)
{{{2sin(x)+cos(x)=1.5}}}<-->{{{cos(x)=1.5-2sin(x)}}}-->{{{(cos(x))^2=2.25-6sin(x)+(2sin(x))^2}}}<-->{{{1-(sin(x))^2=2.25-6sin(x)+4(sin(x))^2}}}
Calling {{{y=sin(x)}}} we can re-write the equation above as
{{{1-y^2=2.25-6y+4y^2}}}<-->{{{5y^2-6y+1.25=0}}}
We solve for {{{y}}} using the quadratic formula:
{{{y=(-(-6) +- sqrt((-6)^2-4*5*1.25))/(2*5)=(6 +- sqrt(36-25))/10=(6 +- sqrt(11))/10}}}
The approximate solutions are
{{{y=(6+sqrt(11))/10=0.9316625}}}
and {{{y=(6-sqrt(11))/10=0.2683375}}} .
With those values of {{{y=sin(x)}}} we set out to find {{{x}}} .
 
For {{{y=sin(x)=0.9316625}}} :
There are two values of {{{x}}} between 0 and {{{2pi}}} (or between {{{0^o}}} and {{{360^o}}} that yield {{{y=sin(x)=0.9316625}}} .
One is in quadrant I and the other in quadrant II.
With {{{2sin(x)=2*0.9316625=1.863325>1.5}}} ,
to get {{{2sin(x)+cos(x)=1.5}}} we need to have {{{cos(x)<0}}} ,
so we are looking foir a quadrant II solution.
{{{highlight(x=1.9426)}}} (in radians, or {{{highlight(111.3^o)}}} in degress) is a solution.
Another possible value is {{{x=1.1990}}} (in radians, {{{68.7^o}}}) ),
but that is in quadrant I, with a positive {{{cos(1.1990)=0.03633}}} , and
{{{2sin(1.199)+cos(1.199)=2*0.9316+0.03633=1.86332+0.03633=2.22665}}}
does not satisfy {{{2sin(x)+cos(x)=1.5}}}.
 
For {{{y=sin(x)=0.2683375}}} :
The solution {{{highlight(x=0.27167)}}} (in radians, or {{{highlight(15.57^o)}}} in degrees), in quadrant I, with {{{cos(x)>0}}} looks promising,
as {{{2sin(x)+cos(x)=2*0.2683375+cos(x)=0.572675+cos(x)}}} can be {{{1.5}}}.
In fact, it can be verified to satisfy {{{2sin(x)+cos(x)= 1.5}}} .{{{2sin(x)+cos(x)=2*0.2683375+cos(x)=0.572675+cos(x)}}}
On the other hand, {{{x=2.9}}} , in quadrant II, while it also yields {{{sin(x)=0.268338}}},
it yields {{{cos(x)=-0.97<0}}} and does not satisfy {{{2sin(x)+cos(x)=1.5}}}.
 
The solutions highlighted above, are the solutions between 0 and {{{2pi}}},
and in general, all solutions can be expressed as
{{{highlight(x=0.27167+2k*pi)}}} or {{{highlight(x=1.9426+2k*pi)}}}
(with {{{highlight(K=integer)}}} and {{{x}}} measured in radians).
The same strategy can be used to solve (b)(i).
(The general solutions I found are
{{{highlight(x=0.8449+2k*pi)}}} or {{{highlight(x=3.5837+2k*pi)}}} ).
 
A COMMON STRATEGY:
The left side of the equation in (a)(ii) is
{{{y=2sin(x)+cos(x)}}}
It is a "linear combination of sine and cosine functions".
That is a periodic function, like sine and cosine.
I can see that its period is {{{2pi}}} .
It must be possible to express it as a single trigonometric function,
maybe {{{y=C*cos(x-D)}}} ,with two constants {{{C}}} and {{{D}}} ,
where the cosine function is shifted right by {{{D}}} and dilated vertically by a factor {{{D}}} .
Now, how could I use trigonometric identities to transform
{{{y=3*cos(x)-4*sin(x)}}} and {{{y=2sin(x)+cos(x)}}}
into a function like {{{y=C*cos(x-D)}}} ?
It required a lot of thinking, and on the Sunday morning after such a Saturday night, I did not trust my brain that much.
I just googled "linear combinations of sine and cosine functions",
and helped myself to someone else's thinking.
Trigonometric identities tell us that
{{{cos(x-D)=cos(D)*cos(x)+sin(D)*sin(x)}}}
so {{{y=C*cos(x-D)=C*(cos(D)*cos(x)+sin(D)*sin(x))=Ccos(D)*cos(x)+Csin(D)*sin(x)}}}
So if a linear combination of sine and cosine functions,
{{{y=A*cos(x)+B*sin(x)}}} is equivalent to {{{y=C*cos(x-D)}}} ,
then {{{Ccos(D)*cos(x)+Csin(D)*sin(x)=A*cos(x)+B*sin(x)}}} for all values of {{{x}}} .
That means that
{{{system(A=Ccos(D),B=CsinD)}}}--->{{{system(tan(D)=B/A,C^2=A^2+B^2)}}}
Although that gives you two choices for C,
it is a formula-driven, apparently less cumbersome, common strategy to solve all four problems.
 
Applying those formulas:
{{{2sin(x)+cos(x)}}} has {{{system(A=1,B=2)}}}--->{{{system(tan(D)=2,C^2=5)}}}
The {{{D}}} angle in quadrant I that has {{{tan(D)=2}}}
measures approximately {{{D=1.10715}}} (in radians).
Using {{{C=sqrt(5)}}} and {{{D=1.10715}}} we would conclude that
{{{2sin(x)+cos(x)=sqrt(5)*cos(x-1.10715)}}}
We re-write the equations that {{{2sin(x)+cos(x)}}} and solve:
(a)(i) {{{2sin(x)+cos(x)=1.5}}}-->{{{sqrt(5)*cos(x-1.10715)=1.5}}}-->{{{cos(x-1.10715)=1.5/sqrt(5)}}}-->{{{cos(x-1.10715)=0.3sqrt(5)}}}
{{{cos(x-1.10715)=0.3sqrt(5)}}}-->{{{system(x-1.10715=0.83548+2k*pi,x-1.10715=-0.83548+2k*pi)}}}-->{{{system(x=0.83548+1.10715+2k*pi,x-1.10715=-0.83548+1.10715+2k*pi)}}}-->{{{highlight(system(x=1.94263+2k*pi,x-1.10715=0.27167+2k*pi))}}}
(a)(ii){{{2sin(x)+cos(x)=0}}}-->{{{sqrt(5)*cos(x-1.10715)=0}}}-->{{{cos(x-1.10715)=0}}}-->{{{cos(x-1.10715)=0}}}
{{{cos(x-1.10715)=0}}}-->{{{system(x-1.10715=pi/2+2k*pi,x-1.10715=-pi/2+2k*pi)}}}-->{{{highlight(x=1.10715+k*pi)}}}
 
{{{3cos(x)-4sin(x)}}} has {{{system(A=3,B=-4)}}}--->{{{system(tan(D)=-4/3,C^2=3^2+4^2=9+16=25=5^2)}}}
The {{{D}}} angle in quadrant IV that has {{{tan(D)=-4/3 is}}}
measures approximately {{{D=-0.92730}}} (in radians).
Using {{{C=5}}} and {{{D=-0.92730}}} we would conclude that
{{{3cos(x)-4sin(x)=5cos(x+0.92730)}}}
We re-write the equations that {{{3cos(x)-4sin(x)}}} and solve:
(b)(i) {{{3cos(x)-4sin(x)+1=0}}}-->{{{5cos(x+0.92730)+1=0}}}-->{{{5cos(x+0.92730)=-1}}}-->{{{cos(x+0.92730)=-1/5}}}-->{{{cos(x+0.92730)=-0.2}}}
-->{{{x+0.92730=2k*pi +- 1.77215}}}-->{{{x=-0.92730 +- 1.77215+2kpi}}}
The two solutions are {{{highlight(x=0.8449+2kpi)}}} and
{{{x=-26994+2kpi}}}, which can be written as {{{x=2pi-26994+2kpi=highlight(3.5837+2kpi)}}} .
(b)(ii) {{{3cos(x)=4sin(x)}}}<-->{{{3cos(x)-4sin(x)=0}}}-->{{{5cos(x+0.92730)=0}}}-->{{{cos(x+0.92730)=0}}}
That means {{{x+0.92730=2k*pi +- pi/2}}}-->{{{x=pi/2-0.9273+k*pi)}}}-->{{{highlight(x=0.6435+k*pi)}}}