Question

Given that 0
12 years ago

Answer 1

Look at the graph of sin4x + sin2x - cosx and see how many times it crosses the x-axis. Then try to find these values by solving the equation

Answer 2

http://www.wolframalpha.com/input/?i=++sin%284+x%29%2B+sin%282+x%29-+cos%28x%29

Answer 3

Use the formula
\[

\sin (p)+\sin (q)=2 \sin
\left(\frac{p}{2}+\frac{q}{
2}\right) \cos
\left(\frac{p}{2}-\frac{q}{
2}\right)
\]
to factor the left hand side

Answer 4

to obtain
\[
\sin (2 x)+\sin (4 x)=2 \sin
(3 x) \cos (x)\\
\text {Now}\\
2 \sin(3 x) \cos (x)\ =\cos(x)\\

2 \sin(3 x) \cos (x)\ -\cos(x)=0\\
\cos(x)(2 \sin(3 x) -1)=0\\
\text { so }
\cos(x) =0\\
2 \sin(3 x) -1=0

\]
It is easy from now on

Answer 5

For cos(x)=0, we get
\[
x=\pm \frac \pi 2
\]

Answer 6

for 2 sin(3x) =1\\
\[
\sin(3x) =\frac 12=\sin\left(\frac \pi 6\right)\\

3 x = \frac \pi 6 + k \pi
\\
3x =\pi -\frac \pi 6 + k \pi

\]
You should be able to finish it now

Answer 7

for cos(x)=0, one has to take
\[
x=\frac \pi 2\\
x= 3 \frac \pi 2
\]