
How do you solve sin(x + pi) = 0.5 between 0 - 2pi? | Socratic
Jul 22, 2015 · How do you solve #sin(x + pi) = 0.5# between #0 < x < 2pi#? Trigonometry Trigonometric Identities and Equations Solving Trigonometric Equations 1 Answer
How do you use a calculator to evaluate #sin(-0.65)#? - Socratic
Apr 1, 2017 · sin x = - 0.65 Calculator gives -> - 0.65 --> 2nd --> sin --> x = - 40^@54 Unit circle gives another x that has the same sin value:
How do you solve sin(x + 40) = 0.7 where 0<x<360? - Socratic
Apr 30, 2015 · The value 0.7 is not one of the remarkable values of the function sinus, so: sin(x+40°)=0.7rArr x+40°=arcsin0.7+k360°rArr x=-40°+arcsin0.7+k360° and x+40°=180° …
How do you solve sin 2x + sin x = 0 over the interval 0 to 2pi?
Apr 7, 2016 · 0, pi, (2pi)/3, (4pi)/3, 2pi sin 2x + sin x = 0 Apply the identity: sin 2x = 2sin x.cos x 2sin x.cos x + sin x = 0. sin x(2cos x + 1) = 0 a. sin x = 0 --> x = 0, x = pi and x = 2pi b/ 2cos x …
If sin=0.4, what is sin(-theta) + csc theta? - Socratic
May 26, 2016 · If sin=0.4, what is #sin(-theta) + csc theta#? Trigonometry Right Triangles Relating Trigonometric Functions.
What is the graph of #y=sin (x/3)#? - Socratic
Oct 6, 2014 · (6pi)/(4)=(3pi)/(2) 0,(3pi)/(2),3pi,(9pi)/2,6pi ->x-values These x values correspond to ... sin(0)=0 sin((pi)/(2))=1 sin(pi)=0 sin((3pi)/2)=-1 sin(2pi)=0 Enter the function using the Y= …
How do you find the value for sin x = 0.723? - Socratic
Oct 18, 2015 · Use calculator. sin x = 0.723 --> arc #x = 46^@30#. Trig unit circle gives another arc x that has the same sin value -->
How do you find the value of #sin arcsin(-0.2)#? - Socratic
Apr 15, 2015 · How do you find the value of #sin arcsin(-0.2)#? Trigonometry Inverse Trigonometric Functions Basic Inverse Trigonometric Functions. 1 Answer
How do you solve the following equation # sin x = 0.25# in
Oct 5, 2016 · x_1=0.25 and x_2 = 2.89 The easiest way to solve this type of problem is to plot the sin function and f(x) = 0.25 on one graph (link to this here) and look for the intersections. The …
How do you find all solutions in [0, 2π): - 1 = 0? | Socratic
Oct 11, 2016 · Solution: In [0,2pi) x=pi/6,x=(7pi)/6 and x=(5pi)/6, x= (11pi)/6 3tan^2x-1=0 or3tan^2x=1 or tan^2x=1/3 or tanx=+- 1/sqrt3.