How do you solve #2sin^2x-5sinx+2=0#?

Question

Charles Autterson · Accepted Answer

#x={pi/6}#

Use the substitution method and then factor

Let #u=sinx#

so, #u^2=sin^2x#

#2color(red)(u^2)-5color(red)(u)+2=0#

Now you can factor

Multiply the coefficient of the first term, #2#, with the last term, #2#.

#2*2=4#

Ask yourself what are the factors of #4# that add up to the coefficient of the middle term, #-5#?

Factors of 4

#(1)(4)# #=> NO#
#color(red)((-1)(-4))# #=> color(red)(YES)#
#(2)(2)# #=> NO#
#(-2)(-2)# #=> NO#

Now place those factors in an order that makes it easy to factor by grouping.

#(2u^2color(red)(-4u))+(color(red)(-1u)+2)=0#

Factor out #2u# from the first grouping.

Factor out #-1# from the second grouping.

#color(red)(2u)(u-2)color(red)(-1)(u-2)=0#

Now you can factor out a grouping, #(u-2)#

#(u-2)(2u-1)=0#

Now use the Zero Property

#u-2=0# and #2u-1=0#

#u=2# and #u=1/2#

Now switch back to #sinx#

#sin x=2# and #sin x =1/2#

#cancel(sin x=2)# is discarded because #sin x# oscillates between -1 and 1.

Falling back to trigonometry #sin x# is in the ratio #y/r#

#y/r=1/2# which means that #x=sqrt3# by using the pythagorean theorem #x=sqrt(2^2-1^2)=sqrt(4-1)=sqrt3# or knowing that we have the special triangle , #30,60,90# which corresponds to the sides #1,sqrt3,2.#

The side of length #1# corresponds to #30# degrees which is also #pi/6#

All of that to say that

#sin (pi/6)=1/2#

#x={pi/6}#

I have tutorials on methods of factoring found here, https://tutor.hix.ai

Abigail Armstrong · Answer

undefined To solve the equation $2\sin^2(x) - 5\sin(x) + 2 = 0$, you can use a substitution method. Let $u = \sin(x)$. Then the equation becomes a quadratic equation in terms of $u$, which you can solve using standard quadratic equation solving techniques. Once you find the solutions for $u$, you can then find the corresponding solutions for $x$ by solving for $x$ using the inverse sine function.