trig problem !

[wallada]

i have a math problem that i do not really understand :

sin²x - 0,64 = 0

the answer in the back of the book says that it's 0.93 .

i dont understand how they got to that answer because i thought that to solve it , you had to do the square root of sin²x and of 0,64 since the are both perfect squares.

 

[tkhunny]

What a strange answer.

Factor:

(sin(x) + 0.8)(sin(x) - 0.8) = 0

Solve sin(x) + 0.8 = 0 -- Lots of solutions.
Solve sin(x) - 0.8 = 0 -- Lots of solutions.

I'm guessing desired slutions are in some very narrow range of values. Otherwise, there are infinitely many.

 

[wallada]

can you explain the equation in more detail please, i am kinda getting confused.

 

[tkhunny]

From Algebra, you should recognize a "Difference of Squares". Do you? If so, so you recall how to factor it? If not, you must review.

 

[soroban]

Hello, wallada!

\(\displaystyle \text{Silve: }\:\sin^2\!x - 0.64 \:=\: 0\)

\(\displaystyle \text{The answer in the back of the book says that it's }0.93\)

We have: .\(\displaystyle \sin^2\!x \:=\:0.64\)

Take square roots: .\(\displaystyle \sin x \:=\:\pm0.8\)

\(\displaystyle \text{Then: }\;x \;=\;\sin^{-1}(0.8) \;=\; 0.927295218\)

\(\displaystyle \text{Therefore: }\;x \;\approx\;\pm0.93 + \pi n\text{ radians.}\)

 

[tkhunny]

Really, though, Wallada, never do that. Use factoring that GIVES you both sets of answers, rather than having to REMEMBER to include both when taking the square root.