Question #82636

2 Answers
Tazwar Sikder
Jul 24, 2017

#1)# #x = frac(pi)(6), pi - frac(pi)(6), pi + frac(pi)(6), 2 pi - frac(pi)(6)#

#2)# #x = frac(pi)(3), frac(2 pi)(3)#

Explanation:

#1)# #2 sin^(2)(x) = frac(1)(2)#

#Rightarrow sin^(2)(x) = frac(1)(4)#

#Rightarrow sin(x) = pm sqrt(frac(1)(4))#

#Rightarrow sin(x) = pm frac(1)(2)#

Let the reference angle be #x = arcsin(frac(1)(2)) Rightarrow x = frac(pi)(6)#.

The value of #sin(x)# are both positive and negative, which means #x# will be in all four quadrants.

#Rightarrow x = frac(pi)(6), pi - frac(pi)(6), pi + frac(pi)(6), 2 pi - frac(pi)(6)#

#therefore x = frac(pi)(6), frac(5 pi)(6), frac(7 pi)(6), frac(11 pi)(6)#

#2)# #3 cos(x) = 2 sin^(2)(x)#

One of the Pythagorean identities is #cos^(2)(x) + sin^(2)(x) = 1#.

We can rearrange it to get:

#Rightarrow sin^(2)(x) = 1 - cos^(2)(x)#

Let's apply this rearranged identity to our proof:

#Rightarrow 3 cos(x) = 2 (1 - cos^(2)(x))#

#Rightarrow 3 cos(x) = 2 - 2 cos^(2)(x)#

#Rightarrow 2 cos^(2)(x) + 3 cos(x) - 2 = 0#

#Rightarrow 2 cos^(2)(x) + 4 cos(x) - cos(x) - 2 = 0#

#Rightarrow 2 cos(x) (cos(x) + 2) - 1 (cos(x) + 2) = 0#

#Rightarrow (cos(x) + 2)(2 cos(x) - 1) = 0#

Using the null factor law:

#Rightarrow cos(x) + 2 = 0 or 2 cos(x) - 1 = 0#

#Rightarrow cos(x) = - 2 or 2 cos(x) = 1#

#Rightarrow cos(x) = - 2 or cos(x) = frac(1)(2)#

However, #cos(x) ne - 2#.

#Rightarrow cos(x) = frac(1)(2)#

Let the reference angle be #x = arccos(frac(1)(2)) Rightarrow x = frac(pi)(3)#.

The value of #cos(x)# is positive, so it will be in the first and second quadrants.

#Rightarrow x = frac(pi)(3), pi - frac(pi)(3)#

#Rightarrow x = frac(pi)(3), frac(2 pi)(3)#

Jim G.
Jul 24, 2017

#30^@" and " 60^@#

Explanation:

#"isolate "sin^2 x" by dividing both sides by 2"#

#rArrsin^2x=1/4#

#color(blue)"take the square root of both sides"#

#sqrt((sin^2x))=+-sqrt(1/4)larr" note plus or minus"#

#rArrsinx=+-1/2#

#"since solution is only required in first quadrant " x<90^@#

#"then solve "sinx=+1/2#

#rArrx=sin^-1(1/2)=30^@#

#(2)#

#"using the "color(blue)"trigonometric identity"#

#•color(white)(x)sin^2x+cos^2x=1#

#rArrsin^2x=1-cos^2x#

#rArr3cosx=2sin^2x#

#rArr3cosx=2(1-cos^2x)#

#rArr3cosx=2-2cos^2x#

#"move all terms to left side and equate to zero"#

#rArr2cos^2x+3cosx-2=0#

#"factorising quadratically gives"#

#(2cosx-1)(cosx+2)=0#

#rArr2cosx-1=0" or " cosx+2=0#

#cosx+2=0rArrcosx=-2larrcolor(red)" not valid"#

#"since "-1<=cosx<=1#

#2cosx-1=0rArrcosx=1/2#

#rArrx=cos^-1(1/2)=60^@larrx<90^@#

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