Verify the identity.

sin2x/cosx + sinx = 2sinx

Solve for all values of x:

6cos2x-7cosx-3=0

If f(x)=cos1/2x-sin2x find the value of
f(pi)

the first identity is not correct

LS = sin2x/cosx + sinx
= 2sinxcosx/cosx + sinx
= 2sinx + sinx
= 3 sinx

You have RS = 2sinx

6cos2x-7cosx-3=0
You probably typed this incorrectly and really meant
6cos^2 x-7cosx-3=0
if so, let cosx = y, then our equation becomes
6y^2 - 7y - 3 = 0
(2y - 3)(3y + 1) = 0
y = 2/3 or y = -1/3

cosx = 2/3 or cosx = -1/3

for cosx = 2/3, x is in I or IV
x = 48.2 or 311.8°

for cosx = -1/3, x is in II or III
x = 109.5 or 250.5°

for the last one, sub in x = π

To verify the identity sin2x/cosx + sinx = 2sinx, we can simplify both sides of the equation and show that they are equal.

Starting with the left-hand side (LHS):
sin2x/cosx + sinx

To combine the fractions, we need a common denominator, which in this case is cosx. So, we have:
(sin2x + cosx * sinx) / cosx

Using the double angle identity sin2x = 2sinx*cosx, we can replace sin2x with its equivalent:
(2sinx*cosx + cosx * sinx) / cosx

Factoring out cosx from the numerator, we get:
(2sinx*cosx + sinx*cosx) / cosx

Combining like terms, we have:
(3sinx*cosx) / cosx

The cosx in the numerator cancels out with the cosx in the denominator, leaving us with:
3sinx

So, the simplified left-hand side (LHS) of the equation is 3sinx.

Now, let's simplify the right-hand side (RHS):
2sinx

Since the left-hand side (LHS) and the right-hand side (RHS) both simplify to 3sinx, we can conclude that the identity sin2x/cosx + sinx = 2sinx is verified.

To solve the equation 6cos2x - 7cosx - 3 = 0, we can use a substitution to simplify it. Let's substitute a new variable u for cosx.

Let u = cosx. Then the equation becomes:
6u^2 - 7u - 3 = 0

Next, we can factor this quadratic equation:
(2u + 1)(3u - 3) = 0

Setting each factor equal to zero, we have:
2u + 1 = 0 or 3u - 3 = 0

Solving each equation for u, we get:
2u = -1 or 3u = 3

Dividing both sides of the equations by 2 and 3 respectively, we have:
u = -1/2 or u = 1

Now, substituting u back with cosx, we get two possible solutions for cosx:
cosx = -1/2 or cosx = 1

Using the unit circle and the periodicity of cosine, we find the solutions for x:
For cosx = -1/2, the values of x are π/3 + 2πn and 5π/3 + 2πn, where n is an integer.
For cosx = 1, the values of x are 2πn, where n is an integer.

Finally, to find the value of f(pi) for the function f(x) = cos(1/2x) - sin(2x), we can simply substitute x = pi into the expression.

f(pi) = cos(1/2 * pi) - sin(2 * pi)

cos(1/2 * pi) = cos(pi/2) = 0
sin(2 * pi) = sin(0) = 0

Substituting these values into the expression, we get:

f(pi) = 0 - 0 = 0

Therefore, the value of f(pi) is 0.