Respuesta :
4. Simplify the expression.
sine of x to the second power minus one divided by cosine of negative x
sin2x+cos2x=1 1−sin2x=cos2x
cos2(x)/(sin(x)−csc(x)) csc(x)=1/sin(x) cos2(x)/(sin(x)− 1/sin(x))= cos2(x)/((sin2(x)− 1)/sin(x)) sin2(x)− 1=-cos2(x) cos2(x)/(( -cos2(x))/sin(x))
=-sin(x)
the answer is the letter a) -sin x
5. Find all solutions in the interval [0, 2π). (6 points)
sin2x + sin x = 0
using a graphical tool
the solutions x1=0 x2=pi x3=3pi/2
the answer is the letter D) x = 0, π, three pi divided by two

Answer with explanation:
Ques 4)
We are given a trignometric expression by:
[tex]\dfrac{\sin^2x-1}{\cos (-x)}[/tex]
which could also be written as:
[tex]=\dfrac{-(1-\sin^2x)}{\cos x}[/tex]
since, we know that:
[tex]\cos (-x)=\cos x[/tex]
Also,
[tex]1-\sin^2 x=\cos^2 x[/tex]
Hence, we get:
[tex]\dfrac{\sin^2x-1}{\cos (-x)}=\dfrac{-\cos^2x}{\cos x}\\\\\\\dfrac{\sin^2x-1}{\cos (-x)}=-\cos x[/tex]
The correct option is:
D) -cos x
Ques 5)
We are asked to find the solution in the interval [0,2π) of the expression:
[tex]\sin 2x+\sin x=0[/tex]
This expression could also be written as:
[tex]2\sin x\cos x+\sin x=0\\\\i.e.\\\\\sin x(2\cos x+1)=0\\\\i.e.\\\\Either\ \sin x=0\ or\ 2\cos x+1=0[/tex]
If
[tex]\sin x=0[/tex]
Then the possible values of x are:
[tex]x=0,\pi[/tex]
and if
[tex]2\cos x+1=0\\\\i.e.\\\\\cosx=\dfrac{-1}{2}[/tex]
We know that the cosine function is negative in second and third quadrant and the possible values where x is negative is:
[tex]x=\dfrac{2\pi}{3}\ ,\ x=\dfrac{4\pi}{3}[/tex]
Hence, all the solutions of the given expression that will lie in the given region is:
[tex]x=0,\ \pi\ ,\ \dfrac{2\pi}{3}\ ,\ \dfrac{4\pi}{3}[/tex]