Higher Maths Homework Help?

3 cos x + 2 = 4

3 cos x = 2

cos x = 2/3

x ≈ 0.841, 5.442

2 sin^2 x = 1

sin^2 x = 1/2

sin x = ± √2/2

x = π/4, 3π/4, 5π/4, π/4

3 tan 2x + 4 = 1

3 tan 2x = -3

tan 2x = -1

2x = 3π/4, 5π/4, 9π/4, 13π/4

x = 3π/8, 5π/8, 9π/8, 13π/8

Hi KillieTillDie7, better recall these again and again

If sin x = sin t

then x = npi + (-1)^n t

And if cos x = cos t

then x = 2npi +/- t

In case of tangent function, if tan x = tan t

then x = npi + t

In all the above cases n can assume any value 0,1,2,3,4,5,6,...............................

Now let us solve the problems given by you

1) 3 cos x + 2 = 4

So cos x = 2/3

x = arc cos 2/3 = 48.2 deg

In radian 48.2/180 * pi = 0.268 pi

Hence x = 2npi +/- 0.268 pi

2) sin x = 1/,/2 = sin pi/4

So x = npi + (-1)^n pi/4

3) tan 2x = -1

or tan 2x = - tan pi/4 = tan (-pi/4)

So 2x = npi + (-pi/4)

Or x = 1/2 ( npi -pi/4) = (4n-1) *pi/8

* 3 cosx+2 =4

3cosx=2

cosx= 2/3

shift cos

x=48 x=360-48

x=312