plzz help urgent, calculus :)

Question

Solve the following two problems related to trigonometric functions and their inverses:

(a) Solve the equation cot x = 2sin(2x) for x E [0,pi}.

(b) Evaluate tan(sin^-1(1/4)).

Answer 1

Evaluate tan(Sin^-1(1/4))

Let θ = Sin^-1(1/4). Then sinθ = 1/4 and 0 < θ < π/2.

Note that the domain of Sin^-1x is [-1,1] and the range is [-π/2, π/2]. If -1 < x < 0, SIn^-1x < 0 and

if 0 < x < 1, Sin^-1x > 0.

cosθ = √(1 - sin²x) = √(15/16) = (√15) / 4

So, tan(Sin^-1(1/4)) = tanθ = sinθ / cosθ = (1/4) / (√15/4) = 1 / √15 = √15/15

Answer 2

a) cotx = 2sin(2x)= 2(2sinxcosx)= 4sinxcosx

= cosx/sinx = 4sinxcosx

1/sinx = 4sinx and cosx=0, x=pi/2 for 0<x<pi

sin^2(x) = 1/4

sinx = + or - 1/2

x = arcsin(+ or -1/2) = pi/6, 5pi/6

x = pi/6, pi/2, 5pi/6

check the answers, plug each of the 3 angles into the original equation

cot30= 2sin60

sqr3 = 2(sqr3/2)= sqr3

cot90 = 2sin180

0 = 2(0) = 0

cot150 = 2sin300

-sqr3 = 2(-sqr3/2) =-sqr3

b) tan(arcsin(1/4) = (sqr15)/15

the angle whose sine = 1/4 has opposite side/hypotenuse = 1/4

opposite side =1, hypotenuse = 4,

and adjacent side = +/-sqr(16-1) =+/-sqr15, using the Pythagorean Theorem

tangent of that same angle = opposite side/adjacent side = +/-1/sqr15 = +/-(sqr15)/15

check the answer, by finding the actual angle, use a calculator with inverse trig functions

arcsin.25 = about 14.4775 degrees

tan14.4775 = about .2582 = about sqr15/15

also the angle could be -14.4775

so tangent of the negative angle = about .2582 = about sqr15/15 too

another way to check the answer in part a) is use a graphing calculator and see where the graph intersects the x axis

with desmos it's about .53 and 2.62, which equals about 30 and 150 degrees = pi/6 and 5pi/6

90 degrees or pi/2 is also a solution, but doesn't show up in the graph as an intercept