I know some of them are similar,but it would be grate to see each of them.Thanks

Question

1)sin(2x)+cosx=0

 

2)cos^2x+sin^2x(x/2)

 

They are both in 0 to 2pi interval .Thanks

Answer 1

For # 1, use the formula for sin 2x to replace sin 2x by 2sinxcosx; then you factor the equation & solve it.

 

your second problem is very peculiar; it's not an equation; the second term begs to be clarified by parenthesization--just what is the argument of sine?  Is x/2 actually a factor in the second term??

Answer 2

To solve trigonometric equations we use identities so that the arguments are the same.

 

Problem 1) First notice that the argument of sine is 2x and cosine is x.  Use the identity sin(2x)=2sin(x)cos(x).

 

2sin(x)cos(x)+cos(x)=0

 

2) Factor the equation.

 

cos(x)(2sin(x)+1)=0

 

3) Use the zero product rule.  That is set each factor = to zero and solve for x.

 

cos(x) = 0     or     2sin(x)+1=0

 

x = cos^-1(0) = pi/2 or 3*pi/2

 

2sin(x) = -1

sin(x) = -1/2

x=sin^-1(-1/2)=7*pi/6 or 11*pi/6

 

x = {pi/2, 3*pi/2, 7*pi/6, 11*pi/6}