Can you please explain why we're dividing by all of that under the radical in the attached pic? I read the chapter (6.2) but saw nothing about that part and I have no idea why what I did (in 2nd image) didn't get the right answer? Thanks! Given equation:V3 sin 0 – cos 0 = 1, where 0 < 0 < 360Calculation:Given trigonometric equation isV3 sin 0 – cos 0 = 1Divide both sides of equation by V(V3) +(-1)2 = 2이(용) -cos 0 (글) %3D글sin 0 cos 30 – cos 0 sin 30° = 31 J3 sino -cosÔ ltros6/-2.3Sint-Sin'o-2Sint3sin o-65int +3(os o1-sin't4sin A-6s10+2-04sin'A-4sinB (sing -)-2(sıno-1)=074sino-2)(Sino-1)=0sing-I4sind-2sind +2=021

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Description: Can you please explain why we're dividing by all of that under the radical in the attached pic? I read the chapter (6.2) but saw nothing about that part and I have no idea why what I did (in 2nd image) didn't get the right answer? Thanks! Given equa

Given equation is 3sinθ-cosθ=13sinθ=1+cosθ Squaring both sides.…

Answered: Can you please explain why we're…

Trigonometry (11th Edition)

11th Edition

ISBN:9780134217437

Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels

Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels

Chapter1: Trigonometric Functions

Section: Chapter Questions

Problem 1RE: 1. Give the measures of the complement and the supplement of an angle measuring 35°.

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Can you please explain why we're dividing by all of that under the radical in the attached pic? I read the chapter (6.2) but saw nothing about that part and I have no idea why what I did (in 2nd image) didn't get the right answer?

Thanks!

Given equation:
V3 sin 0 – cos 0 = 1, where 0 < 0 < 360
Calculation:
Given trigonometric equation is
V3 sin 0 – cos 0 = 1
Divide both sides of equation by V(V3) +(-1)2 = 2
이(용) -
cos 0 (글) %3D글
sin 0 cos 30 – cos 0 sin 30° =

31 J3 sino -cosÔ l
tros6
/-
2.
3Sint-
Sin'o-2Sint
3sin o-65int +3
(os o
1-sin't
4sin A-6s10+2-0
4sin'A-
4sinB (sing -)-2(sıno-1)=0
74sino-2)(Sino-1)=0
sing-I
4sind-2sind +2=0
21

Expert Solution

Step 1

Given equation is

$\sqrt{3} \sin θ - \cos θ = 1 \sqrt{3} \sin θ = 1 + \cos θ$

Squaring both sides.

$3 \sin^{2} θ = 1 + \cos^{2} θ + 2 \cos θ 3 (1 - \cos^{2} θ) = 1 + \cos^{2} θ + 2 \cos θ 4 \cos^{2} θ + 2 \cos θ - 2 = 0 2 \cos^{2} θ + \cos θ - 1 = 0 2 \cos^{2} θ + 2 \cos θ - \cos θ - 1 = 0 2 \cos θ (\cos θ + 1) - (\cos θ + 1) = 0 (\cos θ + 1) (2 \cos θ - 1) = 0 \cos θ = - 1, \cos θ = \frac{1}{2}$

Step 2

Now,

$\cos θ = - 1 \Rightarrow θ = 180 ° \cos θ = \frac{1}{2} \Rightarrow θ = 60 °, 300 °$

But $θ = 300 °$ does not satisfy the equation . Therefore, solutions of given differential equation are

$θ = 60 °, 180 °$

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