Help solve an equation for a trigonometric function

Let x 2=a, cosx (1-sinx)=(cosa 2-sina 2) (1-2sina*cosa)=(cosa 2-sina 2) (cosa 2+sina 2--2sina*cosa)=(cosa+sina)(cosa-sina) [cosa-sina] 2 =(cosa+sina) (cosa-sina) and divide by cosa

The equation = (1+tana) (1-tana)=1 ( 2-1) can be calculated as tana=[(4-)], so a=arctan[(4-)].

With a universal formula.

Solution: Remember t=tan(x 2).

Rule. sinx=2t/(1+t^2)

cosx=(1-t^2)/(1+t^2)

There is substitution in the original form.

1-t 2) (1-t) 2=1 ( 2-1) factorization, the numerator and denominator are approximated.

1+t)/(1-t)=1/(π/2-1)

The solution is t=tan(x 2)=4-).

So x 2=arctan[(4-)].

The first method is to reduce the cosine by equal the sum of one-half. The sum of squares of the sine cosine is equal to 1, and the sine and cosine values are solved in conjunction with each other, and then the sine is divided by the cosine to get the correct value.

The second method is to square the two sides of four primary colors. You can get the results of sine and cosine, and then add 1 pass to form the sum of squares of the cosine and cosine, so that the fractional numerator and denominator are all about the quadratic secondary formula of sine and cosine, and then it can be converted into correct.

The square gives 1 - 2sinacosa 1 4, so sinacosa 3 8 0, and a (0, )sina 0, so cosa 0, since sina - cosa 0, then sina cosa 0, then tana 1, which is obtained by dividing both sides of 8sinacosa 3 3 (sin a cos a) by cos a.

8tana 3 (tan a 1), tana (4 7) 3 .

The question is wrong.,Change option A to positive.,Just choose A.。。

It's the most basic trigonometric identity transformation.

In essence, the universal formula f(tana)=sinacosa=1 2*sin2a=1 2*2tanx [1+(tana) 2].

tanx/[1+(tana)^2]

f( 3) = 3 (1+3) = 3 4, your answer is to convert f(tana)=sinacosa=sina cosa*(cosa) 2=tana*[1 (seca) 2].

tana*[1+(tana) 2], which involves the formula cosa=1 seca, 1+(tana) 2=(seca) 2, is no longer studied in high school.

24cosx+7sinx-3=0, cos(x-t)=3 25, where t=arctan(7 24), so x-t=2k soil arccoa(3 25), k belongs to z, so x=t+2k soil arccoa(3 25), the other problem is not clear coefficient.

The first problem should be the expression of the function finding the zero point of a known quadratic function, where the zero point is 2, 1 and the double zero point 3

So the third problem of the function expression f(x) (x 2)(x 1)(x 3) is to solve the equation.

sin4(x＋8°)＝25/31

4(x＋8°)＝arcsin (25/31)x＝ (25/31)－8°

The second question is what to ask for, which has not yet been written.

If you can't see the question clearly, you may be able to solve it by looking at the textbook and continuing to use your brain, but you can't ask again, right?

I lost points in the first big question, and I was a fart in the test.

Let y2-y0 = δy, x2-x0 = δxRule.

y+rsina)*tana+1 2*(posture closure and δx+rcosa)=0 square virtual range both sides are divided by cosa, and the traces are then.

y+rtana)*tana+1/2*（δx+r)=0r（tana)^2 +δy*tana +1/2（δx+r）=0tana = 2r =

So: a = arc tan