How do I do this problem? (trig)?

Question

5 sinx^2 - 2sinx=cosx^2

I need to see the steps please

Answer 1

5 sin² x - 2sin x = cos² x
5 sin² x - 2sin x = 1 - sin² x
6 sin² x - sin x - 1 = 0
(3sin x + 1)(2sin x - 1) = 0
sin x = -1/3 or sin x = 1/2

assuming 0 ≤ x ≤ 2π,
if sin x = -1/3,
x = 2.8018 or x = 5.9433
if x = 1/2, x = π/6 or x = 5π/6

Answer 2

i'm going to assume your notation means
(sinx)^2 and not sin(x^2)

you should know some of the trig identies
such as sinx^2 + cosx^2 = 1

this will give us

5sinx^2 -2sinx=1-sinx^2
6sinx^2-2sinx-1=0

now you have a quadratic equation
maybe it is easier for you to see if we introduce
an intermediate variable
y = sinx
which gives us
6y^2-2y-1=0
y=[1+/-sqrt(7)]/6
but y is sinx, so

x=arcsin{[1+/-sqrt(7)]/6}

so evaluate. using the calculator utility included with
the Windows operating system, I get
in degrees, approximately
37.4 degrees for the + case
-15.9 degrees for the - case

Answer 3

go and get your calculator first