Solve the following trigonometric equation for 0to2pi
(a) cos(4x) - 9 sin(2x) + 4 = 0

x = pi/12, 5pi/12, 13pi/12, 17pi/12.

Step-by-step explanation:

Note that cos (4x) =  1 - 2sin^2 (2x)

Substituting we have:

1 - 2sin^2 (2x) - 9 sin(2x) + 4 = 0

2 sin^2 (2x) + 9 sin(2x) - 5 = 0

(2 sin 2x - 1 )(sin 2x + 5 ) = 0

sin 2x = 1/2 and sin 2x = -5.

sin 2x = 1/2,   gives 2x = pi/6, 5pi/6, 13pi/6, 17pi/6.

There are no solutions to sin 2x = -5  because the range of sin x  is -1 to +1.

So x = pi/12, 5pi/12, 13pi/12, 17pi/12.

answer: cos [   arcsin (3/5) - arccos (1/2) ] =

step-by-step explanation: × = 3/5 = 0.6--> x = 36.87 and x = 180 - 36.87 = 143.13 deg

cos y = 1/2 --> y = +- 60 deg a. cos (x - y) = cos (143.13 - 60) = cos 83,13  

=

0.12

b. cos (36.87 - 60) = cos (-23.23) = 0.92

38.6°

step-by-step explanation:

resultant is 10 pound

smaller force = 8 pound

angle formed = tan⁻¹(

                        =tan⁻¹(

                          =tan⁻¹(0.8)

                        =38.6°