Solving Trig Equations class Activity 0 2 cos²x-√3 cosx=0 [0, 360) 6 Sinx + Cosx=1 [0,...
4. Solve for x, 0 SX < 360° a) 2 sinx + 3 = 0 b) 7cosx + 2 = 0 c) 2sinx - sinx = 0 d) 2cos2x - cosx - 1 = 0
please solve and explain detailed sinx + cosx= 1 sin (x) + cos(x) = 1
Solve the following equations for x if 0° < 0 < 360°. 36. 2 cos 20 + sin 0 = 1 35. 1 - 4 cos 0 = -2 cos2 37. sin (30 – 45) = -V3 38. cos 30 = -2
0 bie 3x2_1 e CX 3x)2 dx 2 esco cos va va dx 3 l sinx sec? (cosx) dx • 7/2 cosx sin (sinx) dx 10 s S; X=(1+2x9) "dx x2 sixe dx @[Cittam bjpecede 8 4+3x sin (lax)
question c TOPIC 2: Trig Limits TRIG LIMITS: If direct substitution doesn't work, most triglimits can be solved by expressing the limit as a product of these special trig limits. 1. so also Lim 1 0 sinx |-cosx cos X-1 Lim - 0, so Lim =0, BUT Lim so useless! 01-Cos 0 Limsin 0 SQUEEZE THEOREM is covered and used for some proofs in the teaching videos, but will not be tested Q.14 Please evaluate the following limits, without using...
(1 point) If tan x - -1/3, cosx > 0,, then sin 2x- cos 2x - tan 2x - (1 point) Find cos 29 if sin- 13 85
Team Member Team Member Solve by squaring both sides. cot x-cscx sinx + cos x 2 2. Solve by factoring groups 413 sin2x sec x- V3 sec x +2 =8sin2x 4 sin x cos x- 2v3 sin x - 2 cos x + V3 - 0 Solve using identities 3. 1 + tan2x 3 cos(2x) - cos x +1 0 4 tan2x Solve equations in the form A sin(BxC) D = k 4. 250 sin - 125 0 1235 cos...
0 + 360° Square roots of 5(cos 120° + i sin 120°) 0 + 360° (a) Use the formula zk = V(cos 75 (cos 60° + i sin 60°) + i sin to find the indicated roots of the complex number. (Enter your answers in trigonometric form. Let 0 so< 360°.) n n 20 = 21 = V5 (cos 240° + i sin 240°) x (b) Write each of the roots in standard form. Zo = 5 2 + i...
show work 9. tan 37 10. sec 4 11. Find sin(x + y) and cos(x + y) if cosx = - cosy = -— x is in quadrant II and y is in quadrant III. [10] 12. Find the exact value of sin 2x and cos 2x if sin x = and cos x = - [6] 5 13. Simplify tan (x + 3) to a form involving sinx, cosx, and/or tanx. [6]
tion 5 sinx Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function h(x) 3int dt h'(x) = 3 cos(x)In(sinx) - h'(x) = 3 in (cos(x)s in(x))+ 1m3) - h'(x)=3\n(sin(x)cos(x)) - 31nYX h'(x) - 3sin(x) 31n(7) cosx 2 n'(x) = () - 3cos(x) 3inx sinx x h"(x) = 3sin(x)in(cosx) + 3inky 2x