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2

#5sin(x)−2=2sin(x)+4#

#=>3sinx=6#

#=>sinx=2#

#=>x=sin^(-1)2#

But this equation doesn't have any solutions because #sintheta# is defined only within #[-1,1]# hence, #theta# can have a range #[-π/2,π/2]#

So, #sin^(-1)2# is undefined.

Hope this helps.:)

1

Answer:

#"length "~~26.18" m to 2 dec. places"#

Explanation:

#"the length of an arc given it subtends a known angle at"#
#"the centre is"#

#"arc length (l) "="circumference "xx"fraction of circle"#

#rArrl=2pirxx60/360#

#color(white)(l)=2pixx25xx1/6=(25pi)/3~~26.18" m"#

2

Answer:

#"see explanation"#

Explanation:

#"to find some of the ratios we will require the third side of"#
#"the triangle"#

#"since it is right use "color(blue)"Pythagoras' theorem"#

#"third side "=sqrt(11^2-6^2)=sqrt85#

#•color(white)(x)costheta="adjacent"/"hypotenuse"=6/11#

#rArrsectheta=1/costheta=1/(6/11)=11/6#

#•color(white)(x)sintheta="opposite"/"hypotenuse"=sqrt85/11#

#rArrcsctheta=1/sintheta=1/(sqrt85/11)=11/sqrt85#

#•color(white)(x)tantheta=sintheta/costheta#

#color(white)(xxxxxx)=sqrt85/11xx11/6=sqrt85/6#

#rArrcottheta=1/tantheta=1/(sqrt85/6)=6/sqrt85#

2

#(sin^2x+cos^2x+cot^2x)/(1+tan^2x)=cot^2x#

#"LHS"#:

#(color(red)(sin^2x+cos^2x)+cot^2x)/(color(teal)(1+tan^2x))#

We know the #bb"Pythagorean Identities:"#

  • #color(red)(sin^2x+cos^2x=1)#

  • #color(teal)(1+tan^2x=sec^2x)#

  • #color(indigo)(1+cot^2x=csc^2x)#

#bb"Reciprocal Identities:"#

  • #cosx=1/(secx)#

  • #sinx=1/(cscx)#

Employing all of them,

#=(color(red)1+cot^2x)/(color(teal)(sec^2x)#

#=(color(indigo)(1+cot^2x))/(sec^2x)#

#=color(indigo)(csc^2x)/(sec^2x)#

#=((1/sinx)^2)/((1/cosx)^2)#

#=(1/sinx)^2×(cosx/1)^2#

#=cot^2x# #[ because 1/(tanx)=cotx=(cosx)/(sinx)]#

#=> "RHS"#

Hence, proved! :)

Hope this helps. :)

2

Answer:

check the explanation below

Explanation:

#csc(-x)/sec(-x)=(1/sin(-x))/(1/cos(-x))=cos(-x)/sin(-x)# #color(green)(rarr "reciprocal"#

now substitute

#cos(-x)/sin(-x)+cos(-x)/sin(-x)=2cos(-x)/sin(-x)#

#color(blue)(sinx " is odd so "sin(-x)=-sinx"#

#color(blue)( cosx " is even so "cos(-x)=cosx#

Substitute

#=2cosx/-sinx##color(green)(rarreven and odd " functions"#

#=-2cot(x)##color(green)(rarrquoitent#

I hope this was helpful.

1

Answer:

#sinx=3/5#

Explanation:

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

#•color(white)(x)sin^2x+cos^2x=1#

#rArrsinx=+-sqrt(1-cos^2x)#

#"x is in second quadrant where "sinx>0#

#rArrsinx=+sqrt(1-(-4/5)^2)#

#color(white)(rArrsinx)=sqrt(1-16/25)=sqrt(9/25)=3/5=0.6#

0

Answer:

#cos2A=7/25#

Explanation:

Here,

#cosA=4/5 > 0=>I^(st)Quadrant or IV^(th) Quadrant#

We know that,

#cos2A=2cos^2A-1=2(4/5)^2-1#

#:.cos2A=2xx16/25-1=32/25-1#

#=>cos2A=(32-25)/25=7/25 > 0#

#=>cos2A=7/25#

2

#cos^4 x- sin^4 x#

#=(color(red)(cos^2 x+ sin ^2 x))(color(blue)(cos^2 x- sin ^2 x))# #[because a^2-b^2=(a+b)(a-b)]#

We know the Pythagorean Identity,

#color(red)(sin^2x+cos^2x=1)=>color(blue)(sin^2x=1-cos^2x)#

So,

#=color(red)1×(cos^2 x- sin^2 x)#

#=cos^2x-color(blue)((1-cos^2x))#

#=2cos^2x-1#

#=cos2x => "RHS"# #[because "Double Angle Formula:" cos2x=2cos^2x-1=1-2sin^2x=cos^2x-sin^2x]#

Hope this helps. :)

1

Answer:

#(19pi)/6" radians"#

Explanation:

#"to convert from "color(blue)"degrees to radians"#

#color(red)(bar(ul(|color(white)(2/2)color(black)("radian measure "="degree measure "xxpi/180)color(white)(2/2)|)))#

#rArrcancel(570)^(19)/1xxpi/cancel(180)^6=(19pi)/6larrcolor(red)"exact value"#

1

Answer:

# x in { pi/4, pi/2, {3pi}/4, {3pi}/2 } #

Explanation:

#2 sin x cos x = sqrt{2} cos x#

#2 sin x cos x - sqrt{2} cos x = 0#

# cos x(2 sin x - sqrt{2}) = 0#

#cos x = 0# or #sin x = \sqrt{2}/2#

#x = 90^circ + 180^circ k quad# for integer #k# or

#x = 45^circ + 360^circ k# or

# x = 135^circ + 360^circ k#

In radians from #0# to #2pi# that's

# x = { pi/4, pi/2, {3pi}/4, {3pi}/2 } #

Check:

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

# x=pi/4, quad sin(2x)=sin(pi/2)=1 #, #sqrt 2 cos (pi/4) = 1 quad sqrt#

# x=pi/2, quad sin(2x)=0#, #sqrt 2 cos (pi/2) = 0 quad sqrt#

I'll leave the other to you to check.