Question
Solve the equation
x=⎩⎨⎧−arccos(2−1+5)+2kπarccos(2−1+5)+2kπ,k∈Z
Alternative Form
x≈{−51.827292∘+360∘k51.827292∘+360∘k,k∈Z
Alternative Form
x≈{−0.904557+2kπ0.904557+2kπ,k∈Z
Evaluate
sin2(x)=cos(x)
Use sin2(x)=1−cos2(x) to rewrite the expression
1−cos2(x)=cos(x)
Move the expression to the left side
1−cos2(x)−cos(x)=0
Rewrite in standard form
−cos2(x)−cos(x)+1=0
Multiply both sides
cos2(x)+cos(x)−1=0
Substitute a=1,b=1 and c=−1 into the quadratic formula cos(x)=2a−b±b2−4ac
cos(x)=2−1±12−4(−1)
Simplify the expression
More Steps
Evaluate
12−4(−1)
1 raised to any power equals to 1
1−4(−1)
Multiply the numbers
1−(−4)
If a negative sign or a subtraction symbol appears outside parentheses, remove the parentheses and change the sign of every term within the parentheses
1+4
Add the numbers
5
cos(x)=2−1±5
Separate the equation into 2 possible cases
cos(x)=2−1+5cos(x)=2−1−5
Use b−a=−ba=−ba to rewrite the fraction
cos(x)=2−1+5cos(x)=−21+5
Rearrange the terms
cos(x)=2−1+5x∈/R
Calculate
More Steps
Evaluate
cos(x)=2−1+5
Use the inverse trigonometric function
x=arccos(2−1+5)
Calculate
x=−arccos(2−1+5)x=arccos(2−1+5)
Add the period of 2kπ,k∈Z to find all solutions
x=−arccos(2−1+5)+2kπ,k∈Zx=arccos(2−1+5)+2kπ,k∈Z
Find the union
x=⎩⎨⎧−arccos(2−1+5)+2kπarccos(2−1+5)+2kπ,k∈Z
x=⎩⎨⎧−arccos(2−1+5)+2kπarccos(2−1+5)+2kπ,k∈Zx∈/R
Solution
x=⎩⎨⎧−arccos(2−1+5)+2kπarccos(2−1+5)+2kπ,k∈Z
Alternative Form
x≈{−51.827292∘+360∘k51.827292∘+360∘k,k∈Z
Alternative Form
x≈{−0.904557+2kπ0.904557+2kπ,k∈Z
Show Solution
Graph