solve the following trig equation :
7 = 8sin(1/2)x. I can solve this using the half-angle indentity. But i need to solve this using period and phase change equation. how do i do this?
7 = 8sin(1/2)x
sin (1/2)x = .875
so (1/2)x is in the first or second quadrant, and
(1/2)x = 61.04º or (1/2)x = 118.96º
then x = 237.9º or 122.1º
but the period of your curve is 720º
so by adding/subtractin 720º to/from any of the found angles will give you more answers.
in general
x = 122.1 + 720k or
x = 237.9 + 720k , where k is an integer.
To solve the trig equation 7 = 8sin(1/2)x using the period and phase change equation, you'll first need to rewrite it in the form A*sin(Bx + C) + D.
Given the equation 7 = 8sin(1/2)x, we can rewrite it as 7 = 8sin(π/4)(2x), because (1/2)x = π/4 * 2x.
Comparing this with the general form A*sin(Bx + C) + D, we have:
A = 8
B = π/4
C = 0 (no phase shift)
D = 7
Using the period and phase change equation, the period of a general sine function is given by T = 2π / B. In this case, B = π/4, so the period is T = 2π / (π/4) = 8.
To find the phase shift, we use the equation -C/B. Given that C = 0, the phase shift is -0/(π/4) = 0.
Now we can rewrite the equation in the form A*sin(Bx + C) + D:
7 = 8sin(π/4)(2x) becomes 7 = 8sin(2πx/8) + 0.
Therefore, the equation 7 = 8sin(1/2)x can be rewritten in the form 7 = 8sin(2πx/8) + 0.
To solve for x, you can now use algebraic manipulation or graphical methods. Since the amplitude is 8 and the range of the sine function is -1 to 1, it is apparent that there are no solutions for this equation.