Question

Question #7fce1

Answer 1

`x = pi/4 + 2pik, pi/2 + 2pik, 3pi/4 + 2pik` (Period of sin(x) and cos(x) is `2pi`)

Explanation:

Assuming `sin(2x)sin(x) = cos(x)`, begin by using the double angle identity, `sin(2u) = 2sin(u)cos(u)` to rewrite the equation.

`sin(2x)sin(x) = cos(x) rarr 2sin(x)sin(x)cos(x) = cos(x) rarr 2sin^2(x)cos(x) = cos(x)`

Next subtract cos(x) from the right side to the left to make the equation equal to zero.

`2sin^2(x)cos(x)-cos(x) = 0`

Factor out cos(x) and separate

`cos(x)*(2sin^2(x)-1) = 0 rarr cos(x) = 0` and `2sin^2(x)-1 = 0`

Find the exact values using the unit circle or right triangle method, `cos(x) = 0` when `x = pi/2` and `sin(x) = sqrt(2)/2` when `x = pi/4, 3pi/4` (Remember restrictions `sin(x): R(-pi/2, pi/2); cos(x): R(0, pi)`

Answer 2

Answers:
`x = pi/2 + kpi`
`x = pi/4 + (kpi)/2`

Explanation:

`sin 2x.sin x - cos x = 0`
Replace sin 2x by (2sin x.cos x).
`2sin^2 x.cos x - cos x = 0`
`cos x(2sin^2 x - 1) = 0`
Either factor should be zero.
A. cos x = 0. Unit circle give 2 solutions:
`x = pi/2,` and `x = (3pi)/2`
General answer: `x = pi/2 + kpi`
B. `2sin^2 x - 1 = 0` --> `sin^2 x = 1/2`
`sin x = +- sqrt2/2`
a. `sin x = sqrt2/2`. Trig table and unit circle give 2 solutions:
`x = pi/4`, and `x = pi - pi/4 = (3pi)/4`
b. `sin x = -sqrt2/2`. Unit circle gives -->
`x = pi + pi/4 = (5pi)/4` and `x = 2pi - pi/4 = (7pi)/4`
General answer: `x = pi/4 + (kpi)/2`