a. Give the exact solution between 0 and 2pi?

Question

b. Use identities to rewrite the equation in the terms of a single angle.

cos2X + 3sinX = 2

Answer 1

cos 2x+3sinx=2

Cos 2x= 1-2Sin^2x
So (1-2sin^2x)+3sinx=2
1-2sin^2x+3sinx=2
subtract 2 from both sides
-1-2sin^2x+3sinx=0
multiply both sides by -1
1+2sin^2x-3sinx=0
2sin^2x-3sinx+1=0
Factor
(2sinx-1)(sinx-1)=0
Solve each group for x

2sinx-1=0
2sinx=1
sinx=1/2
x=pi/4 or
x=3pi/4

sinx-1=0
sinx=1
x=pi/2

There are 3 solutions.

Answer 2

Use the identity cos(2x) = 1 - 2sin^2(x).

Then you have:

cos(2x) + 3sinx = 1 - 2sin^2(x) + 3sinx = 2
2sin^2(x) - 3sinx + 2 = 0

Factoring

(2sinx -1)(sinx - 1) = 0

2sinx = 1 or sinx = 1/2
sinx = 1

Therefore x = pi/6 and pi/2.

Answer 3

the first one can not receive on [0, 2?) because the arcsine function has variety -?/2 to ?/2. arcsin(-a million/2) = -?/6. it truly is the only and easily one answer. If the question is locate all y on [0, 2?) such that sin(y) = -a million/2, then that is different. The solutions might want to be 7?/6 and 11?/6. those are the quadrant III and IV angles whose reference perspective is ?/6. b) arccos(-a million/2) = 2?/3. lower back, it truly is the only and easily one answer because the arccosine has variety 0 to ?. If the question is to locate all y in [0, 2?) whose cosine is -a million/2, then the solutions are 2?/3 and 4?/3. those are the quadrant II and III angles whose reference perspective is ?/3. c) arctan(a million/2) ? 0.40 six. the fashion of the arctangent function is -?/2 to ?/2. The tangent of ? + arctan(a million/2) is likewise a million/2.

Answer 4

A.
A=cos2x+3sinx=2
cos2x=1-2sin^2x ==> A=1-2sin^2x+3sinx=2 ==> (t=sinx) ==>
1-2t^2+3t-2=0 ==> 2t^2-3t+1=0 ==> Δ=9-8=1 ==> t=2,1 ==>
sinx=1 ==> x=2kpi+pi/2 (pi/2 , 5pi/2 ... )
sinx=1/2 ==> x=2kpi+pi/6 or x= 2kpi+pi-pi/6 ==> x=pi/6,5pi/6
==> x=pi/2,pi/6,5pi/6
B.
1-2sin^2x+3sinx=2

Answer 5

x = pi